Important Questions: Integrals | Mathematics (Maths) Class 12 - JEE PDF Download

Q1: Integrate: ∫ sin³ x cos²x dx
Ans: Given that, ∫ sin³ x cos²x dx
This can be written as:
∫ sin³ x cos²x dx = ∫ sin² x cos²x (sin x) dx
=∫(1 – cos²x ) cos²x (sin x) dx —(1)
Now, substitute t = cos x,
Then dt = -sin x dx
Now, equation can be written as:
Thus, ∫ sin³ x cos²x dx = – ∫ (1-t²)t² dt
Now, multiply t² inside the bracket, we get
= – ∫ (t²-t⁴) dt
Now, integrate the above function:
= – [(t³/3) – (t⁵/5)] + C —(2)
Where C is a constant
Now, substitute t = cos x in (2)
= -(⅓)cos³x +(1/5)cos⁵x + C
Hence, ∫ sin³ x cos²x dx = -(⅓)cos³x +(1/5)cos⁵x + C

Q2: Determine the antiderivative F of “f”, which is defined by f (x) = 4x³ – 6, where F (0) = 3
Ans: Given function: f (x) = 4x³ – 6
Now, integrate the function:
∫4x³ – 6 dx = 4(x⁴/4)-6x + C
∫4x³ – 6 dx = x⁴ – 6x + C
Thus, the antiderivative of the function, F is x⁴ – 6x + C, where C is a constant
Also, given that, F(0) = 3,
Now, substitute x = 0 in the obtained antiderivative function, we get:
(0)4 – 6(0) + C = 3
Therefore, C = 3.
Now, substitute C = 3 in antiderivative function
Hence, the required antiderivative function is x4 – 6x + 3.

Q3: Determine ∫tan⁸x sec⁴ x dx
Ans: Given: ∫tan⁸x sec⁴ x dx
Let I = ∫tan8x sec4 x dx — (1)
Now, split sec4x = (sec²x) (sec²x)
Now, substitute in (1)
I = ∫tan⁸x (sec²x) (sec²x) dx
= ∫tan8x (tan² x +1) (sec²x) dx
It can be written as:
= ∫tan¹⁰x sec² x dx + ∫tan⁸x sec² x dx
Now, integrate the terms with respect to x, we get:
I =( tan¹¹ x /11) + ( tan⁹ x /9) + C
Hence, ∫tan⁸x sec⁴ x dx = ( tan¹¹ x /11) + ( tan⁹ x /9) + C

Q4: Integrate the given function using integration by substitution: 2x sin(x

 + 1) with respect to x.
Ans: Given function: 2x sin(x²+ 1)
We know that, the derivative of x² + 1 is 2x.
Now, use the substitution method, we get
x² + 1 = t, so that 2x dx = dt.
Hence, we get ∫ 2x sin ( x² +1) dx = ∫ sint dt
= – cos t + C
= – cos (x² + 1) + C
Where C is an arbitrary constant
Therefore, the antiderivative of 2x sin(x² + 1) using integration by substitution method is = – cos (x² + 1) + C

Q5: Write the anti-derivative of the following function: 3x² + 4x³
Ans: Given: 3x² + 4x³
The antiderivative of the given function is written as:
∫3x² + 4x³ dx =  3(x³/3) + 4(x⁴/4
= x³ + x⁴
Thus, the antiderivative of 3x²+ 4x³ = x³ + x⁴.

Q6: Evaluate: ∫ 3ax/(b² + c²x²) dx
Ans: To evaluate the integral, I = ∫ 3ax/(b² + c²x²) dx
Let us take v = b² + c²x², then
dv = 2c²x dx
Thus, ∫ 3ax/(b² +c²x²) dx
= (3ax/2c²x)∫dv/v
Now, cancel x on both numerator and denominator, we get
= (3a/2c²)∫dv/v
= (3a/2c²) log |b² +c²x²| + C
Where C is an arbitrary constant.