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**Verify the following :****Q.1. ****Ans.**

L.H.S. =

[Dividing the numerator by the denominator]

[where C_{1} = C – log 2^{2}]

L.H.S. = R.H.S.

Hence proved.**Q.2. ****Ans.**

L.H.S. =

Put x^{2} + 3x = t

∴ (2x + 3) dx = dt

⇒

L.H.S. = R.H.S.

Hence verified.**Evaluate the following:****Q.3.****Ans****.****∴**

Hence, the required solution is**Q.4. ****Ans.**

Let

Hence, the required solution is**Q.5. ****Ans.**

Let** **

Put x + sin x = t ⇒ (1 + cos x) dx = dt

∴

Hence, the required solution is**Q.6.****Ans.**

Let

Hence, the required solution is**Q.7.****Ans.**

Let

Put tan x = t, ∴ sec^{2} x dx = dt

∴

Hence, the required solution is**Q.8. ****Ans.**

Let

= x + C

Hence, the required solution is x + C.**Q.9.****Ans.**

Let

Hence, the required solution is**Q.10.****Ans.**

Put √x = t x = t^{2} ∵ dx = 2t . dt

∴

Hence,**Q.11.****Ans.**

Let

Let I = I_{1} + I_{2}

Now

and

Put a^{2} – x^{2} = t ⇒ – 2x dx = dt

∴

Since I = I_{1} + I_{2}

∴

Hence,[C = C_{1} + C_{2}]**Alternate method:**

Put x = a cos 2θ

∴ dx = a (– 2 sin 2θ) dθ = – 2a sin 2θ dθ

∴

Now x = a cos 2θ

∴

Hence,**Q.12. (Hint : Put x = z ^{4})**

Let

Put x = t^{4} ⇒ dx = 4t^{3} dt

I = I_{1} – I_{2}

Now

Put t^{3} + 1 = z ⇒ 3t^{2} dt = dz

∴

∴ I = I_{1} – I_{2}

Hence,[∵ C = C_{1} - C_{2}]**Q.13.****Ans.**

Let

Put

∴

Hence,**Q.14.****Ans.**

Let

Hence,**Q.15.****Ans.**

Let

[Making perfect square]

Hence,**Q.16.****Ans.**

Let

I = I_{1} – I_{2}

Now

Put x^{2} + 9 = t ⇒ 2x dx = dt

x dx = dt

∴ I = I_{1} – I_{2}

Hence,**Q.17.****Ans.**

Let

(Making perfect square)

Hence,**Q.18.****Ans.**

Let

Put x^{2} = t ⇒ 2x dx = dt ⇒ x dx =

Hence,**Q.19.****Ans.**

Let

Put x^{2} = t for the purpose of partial fractions.

We get

Resolving into partial fractions we put

[where A and B are arbitrary constants]

⇒ t = A + At + B – Bt

Comparing the like terms, we get A – B = 1 and A + B = 0

Solving the above equations, we have

∴(Putting t = x^{2})

Hence,**Q.20.****Ans.**

Let

Hence,**Q.21.****Ans.**

Let

Put x = sin θ ⇒ dx = cos θ dθ

Hence,**Q.22.****Ans.**

Let I =

[∵ 2 cos A cos B = cos (A + B) + cos (A - B)]

Hence,**Q.23.****Ans.**

Let

= tan x – cot x – 3x + C

Hence, I = tan x – cot x – 3x + C.**Q.24.****Ans.**

Let I =

Put

∴

Hence,**Q.25.****Ans.**

Let

= x + 2 sin x + C

Hence, I = x + 2 sin x + C.**Q.26.(Hint : Put x ^{2} = sec θ)**

Let

Put x

∴ 2x dx = sec θ tan θ dθ

∴

So

Hence,

Let

Using the formula,

where h =

Here, a = 0 and b = 2

∴

Here, f(x) = x

f(0) = 0 + 3 = 3

f(0 + h) = (0 + h)

f(0 + 2h) = (0 + 2h)

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

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

Now

Hence,**Q.28.****Ans.**

Let

Here, a = 0 and b = 2

Here f(x) = e^{x}

f(0) = e^{0} = 1

f(0 + h) = e^{0 + h} = e^{h}

f(0 + 2h) = e^{0 + 2h} = e^{2h}

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

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

Hence, I = e^{2 }–1.**Evaluate the following:****Q.29.****Ans.**

Let** **

Put e^{x} = t ⇒ e^{x} dx = dt

Changing the limit, we have

When x = 0 ∴ t = e^{0} = 1

When x = 1 ∴ t = e^{1} = e

Hence,**Q.30.****Ans.**

Let

Put sin^{2} x = t

2 sin x cos x dx = dt

sin x cos x dx =

Changing the limits we get,

When x = 0 ∴ t = sin^{2} 0 = 0; When x =

Hence,**Q.31.****Ans.**

Let

[Making perfect square]

= sin^{– 1} (4 – 3) – sin^{– 1} (2 – 3)

= sin^{– 1} (1) – sin^{– 1} (– 1) = sin^{– 1} (1) + sin^{– 1} (1)

Hence, I = π.**Q.32.****Ans.**

Let

Put 1 + x^{2} = t ⇒ 2x dx = dt ⇒ x dx =

Changing the limits, we have

When x = 0 ∴ t = 1

When x = 1 ∴ t = 2

∴

Hence, I = √2 -1 .**Q.33.****Ans.**

Let...(i)

...(ii)

Adding (i) and (ii) we get,

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

∴**Q.34.(Hint: let x = sinθ)****Ans.**

Let

Put x = sin θ

∴ dx = cos θ dθ

Changing the limits, we get

When x = 0 ∴ sin θ = 0 ∴ θ = 0

Now, dividing the numerator and denominator by cos^{2} θ, we get

Put tan θ = t

∴ sec^{2} θ dθ = dt

Changing the limits, we get

When θ = 0 ∴ t = tan 0 = 0**Long Answer (L.A.)****Q.35. ****Ans.**

Let

Put x^{2} = t for the purpose of partial fraction.

We get

Let

[where A and B are arbitrary constants]

⇒ t = At + 3A + Bt – 4B

Comparing the like terms, we get

A + B = 1 and 3A – 4B = 0

⇒ 3A = 4B

∴

Now

So,

Hence,**Q.36.****Ans.**

Let

Put x^{2} = t for the purpose of partial fraction.

We get

⇒ t = At + Ab^{2} + Bt + Ba^{2}

Comparing the like terms, we get

A + B = 1 and Ab^{2} + Ba^{2} = 0

∴

⇒

So

Hence,**Q.37.****Ans.**

Let...(i)

...(ii)

Adding (i) and (ii), we get

2I = π [0 - (- 1 - 1)] = π(2)

∴ I = π

Hence, I = π.**Q.38.****Ans.**

Let

Resolving into partial fraction, we put

⇒ 2x – 1 = A(x + 2)(x – 3) + B(x – 1)(x – 3) + C(x – 1)(x + 2)

put x = 1, 1 = A(3)(– 2)

put x = – 2, – 5 = B(– 3)(– 5)

put x = 3, 5 = C(2)(5)

Hence,**Q.39.****Ans.**

Let

Put tan^{– 1}x = t

Here f(t) = tan t

∴ f ′(t) = sec^{2} t

= e^{t} . f(t) = e^{t} tan t = - + e ^{tan-1 x} .x + C

Hence, I = - + e ^{tan-1 x} .x + C.**Q.40.(Hint: Put x = a tan ^{2}θ)**

Let

Put x = a tan

dx = 2a tan θ . sec

∴

Hence,

Let

Put

Changing the limits, we have

∴

Hence,

Let

Now, put

∴

Hence,

Let

Put tan x = t

sec

∴

[Dividing the numerator and denominator by t

Put

∴ I = I

Now

Put

Now

Put

So I = I

⇒

Hence,

Let

Dividing the numerator and denominator by cos

Put tan x = t ⇒ sec

Changing the limits, we get

When x = 0, t = tan 0 = 0

∴

Put t

Comparing the coefficients of like terms, we get

a

Now

∴

Hence, I =

Let I =

Hence,

Let I =

log sin (π - x) dx

...(ii)

Adding (i) and (ii), we get

∴...(iii)

...(iv)

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

Put 2x = t ⇒ 2 dx = dt

dx [Changing the limit]

2I = I -π . log 2 [ x]

So

Let I = ...(i)

...(ii)

Adding (i) and (ii), we get

∴

∴cos 2x dx

Put 2x = t

Changing the limits we get

When x = 0 ∴ t = 0; When x =

...(iii)

...(iv)

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

Put 2t = u ⇒ 2 dt = du

∴

[From eq. (ii)]

Hence,**Objective Type Questions****Q.48.is 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 =

∴ I = 2(sin x +cos θ .x)+ C.

Hence, correct option is (a).**Q.49.is equal to ****(a) sin (b – a) log**** (b) cosec (b – a) log****(c) cosec (b – a) log**** (d) sin (b – a) log****Ans.** (c)

Solution.

Let I =

Multiplying and dividing by sin (b – a) we get,

Hence, the correct option is (c).**Q.50. is equal to ****(a) (x + 1) tan ^{ –1} √x – √x + C **

Solution.

Let I =

Put √x = tan θ ⇒ x = tan

∴

Let us take

Put tan θ = t ⇒ sec

∴

∴

∴ 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.is equal to****(a)****(b)****(c)****(d)****Ans.** (a)

Solution.

Let I =

Here f(x) =

Using

∴

Hence, the correct option is (a).**Q.52.is equal to****(a)****(b)****(c)****(d)****Ans.** (d)

Solution.

Let

Put

∴

Hence, the correct option is (d).**Q.53. Ifthen****(a)****(b)****(c)****(d)****Ans.** (c)

Solution.

Let I =

Put

1 = A(x^{2} + 1) + (x + 2) (Bx + C)

1 = Ax^{2} + A + Bx^{2} + Cx + 2Bx + 2C

1 = (A + B)x^{2} + (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 = 2

∴

∴

Putting the given value of I

Hence, the correct option is (c).**Q.54.is equal to****(a)****(b)****(c)****(d)****Ans.** (d)

Solution.

Let I =

∴

Hence, the correct option is (d).**Q.55.is equal to ****(a) log 1 + cos x + C ****(b) log x + sin x + C****(c)****(d)****Ans.** (d)

Solution.

Let I =

∴

Hence, the correct option is (d).**Q.56. Ifthen****(a)****(b)****(c)****(d)****Ans.** (d)

Solution.

Let I =

Put 1 + x^{2} = t ⇒ 2x dx = dt ⇒ x dx =

But I = a(1 + x^{2} )^{3/2} +

Hence, the correct option is (d).**Q.57.is equal to ****(a) 1 ****(b) 2 ****(c) 3 ****(d) 4****Ans.** (a)

Solution.

Let I =

Hence, the correct option is (a).**Q.58.is equal to****(a) 2√2 ****(b) 2( √2 + 1) ****(c) 2 ****(d) 2( √2 - 1)****Ans.** (d)

Solution.

Let I =

Hence, the correct option is (d).**Q.59.is equal to _______.****Ans.**

Let I =

Put sin x = t ⇒ cos x dx = dt

When x = 0 then t = sin 0 = 0; When x =

∴

Hence, I = e – 1.

Q.60.= ________.**Ans.**

Let I =

Put

Let f(x) =

Using

∴

Hence,**Fill in the blanks****Q.61. Ifthen a = ________.****Ans.**

Given that:**Q.62.= ________.****Ans.**

Let I =

Put cos x = t

∴ – sin x dx = dt ⇒ sin x dx = – dt

∴

Hence,**Q.63. The value ofsin ^{3}x cos^{2}x dx is _______.**

Let I =

Let f(x) = sin

= sin

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