AP EAMCET20224 Jul 2022Evening ShiftMathematicsDefinite IntegrationActual
∫ 0 π 4 e tan 2 θ sin 2 θ tan θ d θ =
Options
- A1 2 e 2 - 1
- Be 2 - 1
- Cπ 2
- D2 π 2 - e
Correct answer
A. 1 2 e 2 - 1
Step-by-step solution
I = ∫ 0 π 4 e tan 2 θ sin 2 θ tan θ d θ = ∫ 0 π 4 e tan 2 θ tan 3 θ sec 2 θ 1 + tan 2 θ 2 d θ Let tan θ = t I = ∫ 0 1 e t 2 t 3 1 + t 2 2 d t Let t 2 = u I = ∫ 0 1 e u u 2 1 + u 2 d u = 1 2 ∫ 0 1 e u u 1 1 + u 2 d u By integrating by parts method I = 1 2 e u u - 1 1 + u - ∫ 0 1 e u + e u u - 1 1 + u d u 0 1 = 1 2 e u u - 1 1 + u + ∫ 0 1 e u d u 0 1 = 1 2 - e u u 1 + u + e u 0 1 = 1 2 - e 2 + e - 1 = 1 2 e 2 - 1