NTA Abhyas JEE Main2020MathematicsDefinite IntegrationPractice
The value of the integral I = ∫ 1 3 3 d x 1 + x 2 + x 3 + x 5 is equal to
Options
- Aπ 2
- Bπ 3
- Cπ 12
- Dπ 6
Correct answer
C. π 12
Step-by-step solution
Given integral is I = ∫ 1 3 3 d x 1 + x 2 1 + x 3 Let, tan - 1 ⁡ x = θ ⇒ d x = sec 2 ⁡ θ d θ ∴ I = ∫ π 6 π 3 d θ 1 + tan 3 ⁡ θ = ∫ π 6 π 3 cos 3 ⁡ θ sin 3 ⁡ θ + cos 3 ⁡ θ d θ Applying a + b - x property and adding, we get, 2 I = ∫ π 6 π 3 cos 3 ⁡ θ + sin 3 ⁡ θ sin 3 ⁡ θ + cos 3 ⁡ θ d θ 2 I = θ π 6 π 3 ⇒