Solved Examples - Indefinite Integral | Mock Tests for JEE Main and Advanced 2025 PDF Download

Illustration 1: Integrate the curve x/ (1+x⁴). (1978)
Solution:Let I = ∫xdx / (1 + x⁴)
= 1/2.∫ 2x / (1 + (x²)²) dx
Put x² = u then 2xdx = du
Hence, I = ∫ du/ 2(1 + u²)
= 1/2 tan^-1 u +c
= 1/2 tan^-1(x²)+c

Illustration 2: Integrate sin x. sin 2x.sin 3x + sec²x. cos² 2x + sin⁴ x cos⁴ x. (1979)
Solution:Let I₁ = ∫sin x. sin 2x. sin 3x dx
= 1/4 ∫ sin 4x + sin 2x - sin 6x)dx
= -cos 4x/16 -cos 2x/8 + cos 6x/24
I₂ = ∫sec²x. cos² 2x dx
= ∫ sec²x (2cos²x – 1)2 dx
= ∫ (4cos² x + sec²x- 4)dx
=∫(2cos 2x + sec²x-2)dx
= sin 2x + tan x – 2x
And I₃ = ∫sin⁴x. cos⁴ x dx
= 1/128 ∫(3 - 4cos 4x + cos 8x)dx
= 3x/128 – sin 4x /128 + sin 8x/1024
Hence, I = I₁ + I₂ + I₃
= -cos 4x/16 -cos 2x/8 + cos 6x/24 + sin 2x + tan x – 2x + 3x/128 – sin 4x /128 + sin 8x/1024.

Illustration 3: For any natural umber m, evaluate
∫(x^3m + x^2m + x^m) (2x^2m + 3x^m + 6)^1/mdx, x > 0.(2002)
Solution: For any natural number m, the given integral can be written as
I = ∫(x^3m + x^2m + x^m) [(2x^3m + 3x^2m + 6x^m)^1/m dx] / x
Hence I = ∫(2x^3m + 3x^2m + 6x^m)1/m (x^3m-1 + x^2m-1 + x^m-1)dx
Put (2x^3m + 3x^2m + 6x^m) = t
Then 6x^3m-1 + 6mx^2m-1 + 6mx^m-1)dx = dt
Hence I = ∫t^1/m dt/6m
= [(2x^3m + 3x^2m + 6x^m)^(m+1)/m] / 6(m+1) + c