Test: Evaluating Definite Integrals - JEE MCQ

Let I = ∫(0 to pi/2)x³ sin(tan−1x⁴)/(1+x⁸)dx
Put tan−1 x⁴ = t
4x³/(1+x⁸)dx = dt
⇒x³dx/(1+x⁸) = dt/4
Now, I = 1/4∫sin t dt
=−[1/4 cos t]0 to pi/2 + c
= -1/4[cos(0) - cos(pi/2)] +c
= 1/4[-cos(pi/2) + cos(0)] +c
= ¼[-0+1]
= 1/4 

The given form of integral function (say ∫f(x)) can be transformed into another by changing the independent variable x to t,
Substituting x = g(t) in the function ∫f(x), we get;
dx/dt = g'(t)
or dx = g'(t).dt
Thus, I = ∫f(x).dx = f(g(t)).g'(t).dt
Therefore,  ∫f(g(x)).g'(x).dx  = f(b) - f(a)

 ∫(π/8 to π/4)1/(sin²x cos²x)
= ∫(π/8 to π/4)(sin²x + cos^2x)/(sin²x cos²x) dx
= ∫(π/8 to π/4)(1/cos²x +1/sin²x) dx
= ∫(π/8 to π/4)(sec² x dx + cosec² x dx)
= [tan x - cot x](π/8 to π/4) + c
= [tan(π/8) - tan(π/4)] - [cot((π/8) - cot(π/4)]
= ((2)^½ +1  - 1) - (1/(2)^½ +1 - 1)
= 2

In the question, it should be f’(2) instead of f”(2) 
Explanation:- f(x) = ∫(0 to x) log(1+x²)
f’(x) = 2xdx/(1+x²)
f’(2) = 2(2)/(1+(2)²)
= 4/5

∫(1 to 3) x/(1+x²) dx
Put t = 1+x²
dt = 2x dx
= 1/2∫( 1 to 3) dt/t
= ½  log[t](1 to 3)
= ½[log(1+x²)](1 to 3)
= ½[log 10 - log 2]
= ½ (log(10/2)]
= ½[log 5]
= log(5)^½

Let I=∫(0 to k) 1/[1 + 4x²]dx = π8
Now, ∫(0 to k) 1/[4(1/4 + x²)]dx
= 2/4[tan⁻¹ 2x]0 to k
= 1/2tan^-1 2k − 0 = π/8
1/2tan⁻¹ 2k = π8
⇒ tan⁻¹ 2k = π/4
⇒ 2k = 1
∴ k = 1/2

1 + tan² x = sec² x
∫(π/2 to π/4)(sec² x)/(1 + tan² x)
=> ∫(π/2 to π/4)(sec² x)/(sec² x) dx
= ∫1(π/2 to π/4) dx
= [x](π/2 to π/4)
= π/2 - π/4  =  π/4 

∫x(5^x) dx
u = x,   dv = 5^x dx
du = dx,   v = (5^x/ln 5)
∫x(5^x) dx = [x(5^x)/ln 5](0 to 1) - ∫(0 to 1)(5^x)/ln 5 dx
= 5/ ln 5 - 0 - 1/ln 5[5^x/ln 5](0 to 1)
= 5/ln 5 - 5/(ln 5)² + 1/(ln 5)²
=5/ln 5 - 4/(ln 5)²