NCERT Exemplar-Integrals Notes | EduRev

JEE Revision Notes

JEE : NCERT Exemplar-Integrals Notes | EduRev

The document NCERT Exemplar-Integrals Notes | EduRev is a part of the JEE Course JEE Revision Notes.
All you need of JEE at this link: JEE

Verify the following :
Q.1. NCERT Exemplar-Integrals Notes | EduRev
Ans.
L.H.S. =NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev[Dividing the numerator by the denominator]
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev[where C1 = C – log 22]
L.H.S. = R.H.S.
Hence proved.

Q.2. NCERT Exemplar-Integrals Notes | EduRev
Ans.
L.H.S. =NCERT Exemplar-Integrals Notes | EduRev
Put x2 + 3x = t
∴ (2x + 3) dx = dt
NCERT Exemplar-Integrals Notes | EduRev
L.H.S. = R.H.S.
Hence verified.

Evaluate the following:
Q.3.NCERT Exemplar-Integrals Notes | EduRev
Ans.
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev

NCERT Exemplar-Integrals Notes | EduRev
Hence, the required solution isNCERT Exemplar-Integrals Notes | EduRev

Q.4. NCERT Exemplar-Integrals Notes | EduRev
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence, the required solution isNCERT Exemplar-Integrals Notes | EduRev

Q.5. NCERT Exemplar-Integrals Notes | EduRev
Ans.
Let NCERT Exemplar-Integrals Notes | EduRev
Put x + sin x = t ⇒ (1 + cos x) dx = dt
NCERT Exemplar-Integrals Notes | EduRev
Hence, the required solution isNCERT Exemplar-Integrals Notes | EduRev

Q.6.NCERT Exemplar-Integrals Notes | EduRev
Ans.
LetNCERT Exemplar-Integrals Notes | EduRevNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence, the required solution isNCERT Exemplar-Integrals Notes | EduRev

Q.7.NCERT Exemplar-Integrals Notes | EduRev
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Put tan x = t, ∴ sec2 x dx = dt
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence, the required solution isNCERT Exemplar-Integrals Notes | EduRev

Q.8. NCERT Exemplar-Integrals Notes | EduRev
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
= x + C
Hence, the required solution is x + C.

Q.9.NCERT Exemplar-Integrals Notes | EduRev
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence, the required solution isNCERT Exemplar-Integrals Notes | EduRev

Q.10.NCERT Exemplar-Integrals Notes | EduRevNCERT Exemplar-Integrals Notes | EduRev
Ans.
NCERT Exemplar-Integrals Notes | EduRev
Put √x = t  x = t2 ∵ dx = 2t . dt
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence,NCERT Exemplar-Integrals Notes | EduRev

Q.11.NCERT Exemplar-Integrals Notes | EduRev
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Let I = I1 + I2
NowNCERT Exemplar-Integrals Notes | EduRev
andNCERT Exemplar-Integrals Notes | EduRev
Put a2 – x2 = t ⇒ – 2x dx = dt
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Since I = I1 + I2
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence,NCERT Exemplar-Integrals Notes | EduRev[C = C1 + C2]
Alternate method:
NCERT Exemplar-Integrals Notes | EduRev
Put x = a cos 2θ
∴ dx = a (– 2 sin 2θ) dθ =  – 2a sin 2θ dθ
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Now x = a cos 2θ
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRevNCERT Exemplar-Integrals Notes | EduRev
Hence,NCERT Exemplar-Integrals Notes | EduRev

Q.12. NCERT Exemplar-Integrals Notes | EduRev(Hint : Put x = z4)
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev

NCERT Exemplar-Integrals Notes | EduRev
Put x = t4 ⇒ dx = 4t3 dt
NCERT Exemplar-Integrals Notes | EduRev

I = I1 – I2

Now
NCERT Exemplar-Integrals Notes | EduRev
Put t3 + 1 = z ⇒ 3t2 dt = dz
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
∴ I = I1 – I2
NCERT Exemplar-Integrals Notes | EduRev
Hence,NCERT Exemplar-Integrals Notes | EduRev[∵ C = C1 - C2]

Q.13.NCERT Exemplar-Integrals Notes | EduRev
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev
PutNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence,NCERT Exemplar-Integrals Notes | EduRev

Q.14.NCERT Exemplar-Integrals Notes | EduRev
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence,NCERT Exemplar-Integrals Notes | EduRev

Q.15.NCERT Exemplar-Integrals Notes | EduRev
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev[Making perfect square]
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence,NCERT Exemplar-Integrals Notes | EduRev

Q.16.NCERT Exemplar-Integrals Notes | EduRev
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev

I = I1 – I2

NowNCERT Exemplar-Integrals Notes | EduRev
Put x2 + 9 = t ⇒ 2x dx = dt
x dx = dt
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
∴ I = I1 – I2
NCERT Exemplar-Integrals Notes | EduRev
Hence,NCERT Exemplar-Integrals Notes | EduRev

Q.17.NCERT Exemplar-Integrals Notes | EduRev
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev(Making perfect square)
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence,
NCERT Exemplar-Integrals Notes | EduRev

Q.18.NCERT Exemplar-Integrals Notes | EduRev
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev
Put x2 = t ⇒ 2x dx = dt ⇒ x dx =NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence,NCERT Exemplar-Integrals Notes | EduRev

Q.19.NCERT Exemplar-Integrals Notes | EduRev
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev

Put x2 = t for the purpose of partial fractions.

We getNCERT Exemplar-Integrals Notes | EduRev
Resolving into partial fractions we put
NCERT Exemplar-Integrals Notes | EduRev
[where A and B are arbitrary constants]
NCERT Exemplar-Integrals Notes | EduRev
⇒ t = A + At + B – Bt
Comparing the like terms, we get A – B = 1 and A + B = 0
Solving the above equations, we haveNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev(Putting t = x2)
NCERT Exemplar-Integrals Notes | EduRev
Hence,NCERT Exemplar-Integrals Notes | EduRev

Q.20.NCERT Exemplar-Integrals Notes | EduRev
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence,NCERT Exemplar-Integrals Notes | EduRev

Q.21.NCERT Exemplar-Integrals Notes | EduRev
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev
Put x = sin θ ⇒ dx  = cos θ dθ
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence,NCERT Exemplar-Integrals Notes | EduRev

Q.22.NCERT Exemplar-Integrals Notes | EduRev
Ans.
Let I =NCERT Exemplar-Integrals Notes | EduRevNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
[∵ 2 cos A cos B = cos (A + B) + cos (A - B)]
NCERT Exemplar-Integrals Notes | EduRev
Hence,NCERT Exemplar-Integrals Notes | EduRev

Q.23.NCERT Exemplar-Integrals Notes | EduRev
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
= tan x – cot x – 3x + C
Hence, I = tan x – cot x – 3x + C.

Q.24.NCERT Exemplar-Integrals Notes | EduRev
Ans.
Let I =NCERT Exemplar-Integrals Notes | EduRev
PutNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence,NCERT Exemplar-Integrals Notes | EduRev

Q.25.NCERT Exemplar-Integrals Notes | EduRev
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
= x + 2 sin x + C
Hence, I = x + 2 sin x + C.

Q.26.NCERT Exemplar-Integrals Notes | EduRev(Hint : Put x2 = sec θ)
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev
Put x2 = sec θ
∴ 2x dx = sec θ tan θ dθ
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
SoNCERT Exemplar-Integrals Notes | EduRev
Hence,NCERT Exemplar-Integrals Notes | EduRev

Evaluate the following as limit of sums:
Q.27.NCERT Exemplar-Integrals Notes | EduRev
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev
Using the formula,
NCERT Exemplar-Integrals Notes | EduRev
where h =NCERT Exemplar-Integrals Notes | EduRev
Here, a  = 0 and b = 2
NCERT Exemplar-Integrals Notes | EduRev
Here, f(x) = x2 + 3
f(0) = 0 + 3 = 3
f(0 + h) = (0 + h)2 + 3 = h2 + 3
f(0 + 2h) = (0 + 2h)2 + 3 = 4h2 + 3

..............................

..............................
NCERT Exemplar-Integrals Notes | EduRev
Now
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRevNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence,NCERT Exemplar-Integrals Notes | EduRev

Q.28.NCERT Exemplar-Integrals Notes | EduRev
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev
Here, a  = 0 and b = 2NCERT Exemplar-Integrals Notes | EduRev

Here f(x) = ex 
f(0) = e0 = 1
f(0 + h) = e0 + h = eh 
f(0 + 2h) = e0 + 2h = e2h 
................................

................................
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence, I = e–1.

Evaluate the following:
Q.29.NCERT Exemplar-Integrals Notes | EduRev
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev 
NCERT Exemplar-Integrals Notes | EduRev
Put ex = t ⇒ ex dx = dt
Changing the limit, we have
When x  = 0 ∴ t = e0 = 1
When x  = 1 ∴ t = e1 = e
NCERT Exemplar-Integrals Notes | EduRev
Hence,NCERT Exemplar-Integrals Notes | EduRev

Q.30.NCERT Exemplar-Integrals Notes | EduRev
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Put sin2 x = t
2 sin x cos x dx = dt
sin x cos x dx =NCERT Exemplar-Integrals Notes | EduRev
Changing the limits we get,
When x  = 0 ∴ t = sin2 0 = 0;  When x =NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence,NCERT Exemplar-Integrals Notes | EduRev

Q.31.NCERT Exemplar-Integrals Notes | EduRev
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev[Making perfect square]
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev = sin– 1 (4 – 3) – sin– 1 (2 – 3)
= sin– 1 (1) – sin– 1 (– 1) = sin– 1 (1) + sin– 1 (1)
NCERT Exemplar-Integrals Notes | EduRev
Hence, I = π.

Q.32.NCERT Exemplar-Integrals Notes | EduRev
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev
Put 1 + x2 = t ⇒ 2x dx = dt ⇒ x dx =NCERT Exemplar-Integrals Notes | EduRev
Changing the limits, we have
When x  = 0 ∴ t = 1
When x  = 1 ∴ t = 2
NCERT Exemplar-Integrals Notes | EduRev
Hence, I = √2 -1 .

Q.33.NCERT Exemplar-Integrals Notes | EduRev
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev...(i)
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev...(ii)
Adding (i) and (ii) we get,
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Put cos x = t ⇒ – sin x dx = dt ⇒ sin x dx = – dt
Changing the limits, we have
When x  = 0, t = cos 0 = 1; When x = p, t = cos p = – 1
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRevNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev

Q.34.NCERT Exemplar-Integrals Notes | EduRev(Hint: let x = sinθ)
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev
Put x = sin θ
∴ dx = cos θ dθ
Changing the limits, we get
When x  = 0 ∴ sin θ = 0 ∴ θ = 0
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Now, dividing the numerator and denominator by cos2 θ, we get
NCERT Exemplar-Integrals Notes | EduRev
Put tan θ = t
∴ sec2 θ dθ = dt
Changing the limits, we get
When θ = 0 ∴ t = tan 0 = 0
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev

Long Answer (L.A.)
Q.35. NCERT Exemplar-Integrals Notes | EduRev
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev

Put x2 = t for the purpose of partial fraction.

We getNCERT Exemplar-Integrals Notes | EduRev
LetNCERT Exemplar-Integrals Notes | EduRev
[where A and B are arbitrary constants]
NCERT Exemplar-Integrals Notes | EduRev

⇒ t = At + 3A + Bt – 4B
Comparing the like terms, we get
A + B = 1 and 3A – 4B = 0
⇒ 3A = 4B
NCERT Exemplar-Integrals Notes | EduRev
NowNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
So,NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence,NCERT Exemplar-Integrals Notes | EduRev

Q.36.NCERT Exemplar-Integrals Notes | EduRev
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev

Put x2 = t for the purpose of partial fraction.

We getNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev 
⇒ t = At + Ab2 + Bt + Ba2
Comparing the like terms, we get
A + B = 1 and Ab2 + Ba2 = 0
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
SoNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence,NCERT Exemplar-Integrals Notes | EduRev

Q.37.NCERT Exemplar-Integrals Notes | EduRev
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev...(i)
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev...(ii)
Adding (i) and (ii), we get
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
2I = π [0 - (- 1 - 1)] = π(2)
∴ I = π
Hence, I = π.

Q.38.NCERT Exemplar-Integrals Notes | EduRev
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev
Resolving into partial fraction, we put
NCERT Exemplar-Integrals Notes | EduRev
 2x – 1 = A(x + 2)(x – 3) + B(x – 1)(x – 3) + C(x – 1)(x + 2)
put x  = 1, 1 = A(3)(– 2)NCERT Exemplar-Integrals Notes | EduRev
put x  = – 2, – 5 = B(– 3)(– 5)NCERT Exemplar-Integrals Notes | EduRev
put x  = 3, 5 = C(2)(5)NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence,NCERT Exemplar-Integrals Notes | EduRev

Q.39.NCERT Exemplar-Integrals Notes | EduRev
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev
Put tan– 1x = tNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev

Here f(t) = tan t
∴ f ′(t) = sec2 t
= et . f(t) = et tan t = - + e tan-1 x .x + C
NCERT Exemplar-Integrals Notes | EduRev 
Hence, I = - + e tan-1 x .x + C.

Q.40.NCERT Exemplar-Integrals Notes | EduRev(Hint: Put x = a tan2θ)
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev
Put x = a tan2 θ
dx = 2a tan θ . sec2 θ . dθ
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence,NCERT Exemplar-Integrals Notes | EduRev

Q.41.NCERT Exemplar-Integrals Notes | EduRev
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
PutNCERT Exemplar-Integrals Notes | EduRev
Changing the limits, we have
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence,NCERT Exemplar-Integrals Notes | EduRev

Q.42.NCERT Exemplar-Integrals Notes | EduRev
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Now, put
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence,NCERT Exemplar-Integrals Notes | EduRev

Q.43.NCERT Exemplar-Integrals Notes | EduRev(Hint: Put tanx = t2)
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev
Put tan x = t2
sec2 x dx = 2t dtNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
[Dividing the numerator and denominator by t2]
NCERT Exemplar-Integrals Notes | EduRev
PutNCERT Exemplar-Integrals Notes | EduRev
∴ I = I1 + I2 ...(i)
NowNCERT Exemplar-Integrals Notes | EduRev
PutNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NowNCERT Exemplar-Integrals Notes | EduRev
PutNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
So I = I1 + I2
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence,NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev

Q.44.NCERT Exemplar-Integrals Notes | EduRev
Ans.
LetNCERT Exemplar-Integrals Notes | EduRev
Dividing the numerator and denominator by cos4 x, we have
NCERT Exemplar-Integrals Notes | EduRev
Put tan x = t ⇒ sec2 x dx = dt
Changing the limits, we get
When x  = 0, t = tan 0 = 0
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Put t 2 = u only for the purpose of partial fraction
NCERT Exemplar-Integrals Notes | EduRev
Comparing the coefficients of like terms, we get
a2A + B = 1 and b2A = 1
NowNCERT Exemplar-Integrals Notes | EduRevNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence, I =NCERT Exemplar-Integrals Notes | EduRev

Q.45.NCERT Exemplar-Integrals Notes | EduRev
Ans.
Let I =NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev 
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence,NCERT Exemplar-Integrals Notes | EduRev

Q.46.NCERT Exemplar-Integrals Notes | EduRev
Ans.
Let I =NCERT Exemplar-Integrals Notes | EduRev 
NCERT Exemplar-Integrals Notes | EduRevlog sin (π - x) dx
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev...(ii)
Adding (i) and (ii), we get
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRevNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev...(iii)
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev...(iv)
On adding (iii) and (iv), we get
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Put 2x = t ⇒ 2 dx = dtNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRevdx [Changing the limit]
2I = I -π . log 2 [ x]0π/2 [from eqn. (iii)]
NCERT Exemplar-Integrals Notes | EduRev
SoNCERT Exemplar-Integrals Notes | EduRev

Q.47.NCERT Exemplar-Integrals Notes | EduRev
Ans.
Let I =NCERT Exemplar-Integrals Notes | EduRev ...(i)
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev...(ii)
Adding (i) and (ii), we get
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRevcos 2x dx
Put 2x = tNCERT Exemplar-Integrals Notes | EduRev
Changing the limits we get
When x = 0 ∴ t = 0; When x =NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev...(iii)
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev...(iv)

On adding (iii) and (iv), we get,

NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Put 2t = u ⇒ 2 dt = duNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev[From eq. (ii)]
NCERT Exemplar-Integrals Notes | EduRev
Hence,NCERT Exemplar-Integrals Notes | EduRev

Objective Type Questions
Q.48.NCERT Exemplar-Integrals Notes | EduRevis equal to 
(a) 2(sinx + xcosθ) + C 
(b) 2(sinx – xcosθ) + C 
(c) 2(sinx + 2xcosθ) + C 
(d) 2(sinx – 2x cosθ) + C
Ans. (a)
Solution.

Let I =NCERT Exemplar-Integrals Notes | EduRev 
 NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev

∴ I = 2(sin x +cos θ .x)+ C.

Hence, correct option is (a).

Q.49.NCERT Exemplar-Integrals Notes | EduRevis equal to 
(a) sin (b – a) logNCERT Exemplar-Integrals Notes | EduRev
 (b) cosec (b – a) logNCERT Exemplar-Integrals Notes | EduRev
(c) cosec (b – a) logNCERT Exemplar-Integrals Notes | EduRev
 (d) sin (b – a) logNCERT Exemplar-Integrals Notes | EduRev
Ans. (c)
Solution.

Let I =NCERT Exemplar-Integrals Notes | EduRev 
Multiplying and dividing by sin (b – a) we get,
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence, the correct option is (c).

Q.50.NCERT Exemplar-Integrals Notes | EduRev is equal to 
(a) (x + 1) tan –1 √x – √x + C 
(b) x tan –1 √x – √x + C 
(c) √x – x tan –1 √x + C 
(d) √x – ( x + 1) tan –1 √x + C
Ans. (a)
Solution.

 Let I =NCERT Exemplar-Integrals Notes | EduRev 
Put √x = tan θ ⇒  x = tan2 θ ⇒ dx = 2 tan θ sec2 θ dθ
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Let us take
NCERT Exemplar-Integrals Notes | EduRev
Put tan θ = t ⇒ sec2 θ dθ = dt
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev

∴ I = tan - 1 √x .x - √x+ tan-1 √x + C =

( x + 1) tan - 1 √x -√x+ C
Hence, the correct option is (a).

Q.51.NCERT Exemplar-Integrals Notes | EduRevis equal to
(a)NCERT Exemplar-Integrals Notes | EduRev
(b)NCERT Exemplar-Integrals Notes | EduRev
(c)NCERT Exemplar-Integrals Notes | EduRev
(d)NCERT Exemplar-Integrals Notes | EduRev
Ans. (a)
Solution.

Let I =NCERT Exemplar-Integrals Notes | EduRev 
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Here f(x) =NCERT Exemplar-Integrals Notes | EduRev
UsingNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence, the correct option is (a).

Q.52.NCERT Exemplar-Integrals Notes | EduRevis equal to
(a)NCERT Exemplar-Integrals Notes | EduRev
(b)NCERT Exemplar-Integrals Notes | EduRev
(c)NCERT Exemplar-Integrals Notes | EduRev
(d)NCERT Exemplar-Integrals Notes | EduRev
Ans. (d)
Solution.

LetNCERT Exemplar-Integrals Notes | EduRev
PutNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence, the correct option is (d).

Q.53. IfNCERT Exemplar-Integrals Notes | EduRevthen
(a)NCERT Exemplar-Integrals Notes | EduRev
(b)NCERT Exemplar-Integrals Notes | EduRev
(c)NCERT Exemplar-Integrals Notes | EduRev
(d)NCERT Exemplar-Integrals Notes | EduRev
Ans. (c)
Solution.

Let I =NCERT Exemplar-Integrals Notes | EduRev
PutNCERT Exemplar-Integrals Notes | EduRev
1 = A(x2 + 1) + (x + 2) (Bx + C)
1 = Ax2 + A + Bx2 + Cx + 2Bx + 2C
1 = (A + B)x2 + (C + 2B)x + (A + 2C)
Comparing the like terms, we have
A + B = 0 ...(i)
2B + C = 0 ...(ii)
A + 2C = 1 ...(iii)
Subtracting (i) from (iii) we get      
2C  B = 1 ∴ B = 2C – 1

Putting the value of B in eqn. (ii) we have
2(2C – 1) + C = 0 ⇒ 4C – 2 + C = 0

5C = 2NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Putting the given value of I
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence, the correct option is (c).

Q.54.NCERT Exemplar-Integrals Notes | EduRevis equal to
(a)NCERT Exemplar-Integrals Notes | EduRev
(b)NCERT Exemplar-Integrals Notes | EduRev
(c)NCERT Exemplar-Integrals Notes | EduRev
(d)NCERT Exemplar-Integrals Notes | EduRev
Ans. (d)
Solution.

Let I =NCERT Exemplar-Integrals Notes | EduRev 
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence, the correct option is (d).

Q.55.NCERT Exemplar-Integrals Notes | EduRevis equal to 
(a) log 1 + cos x + C 
(b) log x + sin x + C
(c)NCERT Exemplar-Integrals Notes | EduRev
(d)NCERT Exemplar-Integrals Notes | EduRev
Ans. (d)
Solution.

Let I =NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence, the correct option is (d).

Q.56. IfNCERT Exemplar-Integrals Notes | EduRevthen
(a)NCERT Exemplar-Integrals Notes | EduRev
(b)NCERT Exemplar-Integrals Notes | EduRev
(c)NCERT Exemplar-Integrals Notes | EduRev
(d)NCERT Exemplar-Integrals Notes | EduRev
Ans. (d)
Solution.

Let I =NCERT Exemplar-Integrals Notes | EduRev
Put 1 + x2 = t ⇒ 2x dx = dt ⇒ x dx =NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
But I = a(1 + x2 )3/2 +NCERT Exemplar-Integrals Notes | EduRev
Hence, the correct option is (d).

Q.57.NCERT Exemplar-Integrals Notes | EduRevis equal to 
(a) 1 
(b) 2 
(c) 3 
(d) 4
Ans. (a)
Solution.

Let I =NCERT Exemplar-Integrals Notes | EduRev 
NCERT Exemplar-Integrals Notes | EduRev
Hence, the correct option is (a).

Q.58.NCERT Exemplar-Integrals Notes | EduRevis equal to
(a) 2√2 
(b) 2( √2 + 1) 
(c) 2 
(d) 2( √2 - 1)
Ans. (d)
Solution.

Let I =NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence, the correct option is (d).

Q.59.NCERT Exemplar-Integrals Notes | EduRevis equal to _______.
Ans.
Let I =NCERT Exemplar-Integrals Notes | EduRev 
Put sin x = t ⇒ cos x dx = dt
When x  = 0 then t = sin 0 = 0;  When x =NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRevNCERT Exemplar-Integrals Notes | EduRev
Hence, I = e – 1.

Q.60.NCERT Exemplar-Integrals Notes | EduRev= ________.

Ans.
Let I =NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
PutNCERT Exemplar-Integrals Notes | EduRev
Let f(x) =NCERT Exemplar-Integrals Notes | EduRev
UsingNCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence,NCERT Exemplar-Integrals Notes | EduRev

Fill in the blanks
Q.61. IfNCERT Exemplar-Integrals Notes | EduRevthen a = ________.
Ans.
Given that:NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev

Q.62.NCERT Exemplar-Integrals Notes | EduRev= ________.
Ans.
Let I =NCERT Exemplar-Integrals Notes | EduRev
Put cos x = t
∴ – sin x dx = dt ⇒ sin x dx = – dt
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
NCERT Exemplar-Integrals Notes | EduRev
Hence,NCERT Exemplar-Integrals Notes | EduRev

Q.63. The value ofNCERT Exemplar-Integrals Notes | EduRevsin3x cos2x dx is _______.
Ans.
Let I =NCERT Exemplar-Integrals Notes | EduRev
Let f(x) = sin3 x cos2 x f(– x)
= sin3(– x).cos2 (– x) = – sin3 x cos2 x = – f(x)
NCERT Exemplar-Integrals Notes | EduRev

Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Complete Syllabus of JEE

JEE

Dynamic Test

Content Category

Related Searches

Exam

,

past year papers

,

Viva Questions

,

mock tests for examination

,

NCERT Exemplar-Integrals Notes | EduRev

,

Sample Paper

,

Free

,

NCERT Exemplar-Integrals Notes | EduRev

,

Important questions

,

MCQs

,

Previous Year Questions with Solutions

,

NCERT Exemplar-Integrals Notes | EduRev

,

pdf

,

study material

,

shortcuts and tricks

,

practice quizzes

,

video lectures

,

Summary

,

ppt

,

Extra Questions

,

Objective type Questions

,

Semester Notes

;