₀^{π/4} e^{²θ} ²θ θ dθ =

Correct Answer is : 12(e2−1)1/2 ≤ft( e/2 - 1 )21(2e−1)

Correct Answer is : 12(e2−1)1/2 ≤ft( e/2 - 1 )21(2e−1)

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AP EAPCET 04-Jul-2022 Shift 2 Paper

Section: Mathematics

Question : 75 of 160

Marks: +1, -0

\int\limits_{0}^{\frac{\pi}{4}} e^{\tan^{2}\theta} \sin^{2}\theta \tan\theta d\theta =

Solution: 👈: Video Solution

\text{ (a) Let } I = \int e^{\tan^{2}\theta} \sin^{2}\theta \tan\theta d\theta

= \int e^{\tan^{2}\theta} \times \sin^{2}\theta \frac{\tan\theta \sec^{2}\theta}{1+\tan^{2}\theta} d\theta

= \frac{1}{2} \int 2 e^{\tan^{2}\theta} \times \frac{\tan^{2}\theta}{1+\tan^{2}\theta} \times \frac{\tan\theta \sec^{2}\theta}{1+\tan^{2}\theta} d\theta

Let

\tan^{2}\theta = t

\Rightarrow 2 \tan\theta \sec^{2}\theta d\theta = dt

When

\theta = 0

, then

t = 0

and when

\theta = \frac{\pi}{4}

, then

t = 1

\therefore I = \frac{1}{2} \int\limits_{0}^{1} e^{t} \times \frac{t}{(1+t)^{2}} dt = \frac{1}{2} \int\limits_{0}^{1} e^{t} \left[ \frac{1+t-1}{(1+t)^{2}} \right] dt

= \frac{1}{2} \int\limits_{0}^{1} \cdot e^{t} \left[ \frac{1}{1+t} + \cdot \left( \frac{-1}{(1+t)^{2}} \right) \right] dt

= \frac{1}{2} \left[ \frac{e^{t}}{1+t} \right]_{0}^{1} \qquad \left[ \because \int e^{x} (f(x)+f'(x) dx) = e^{x} f(x)+C \right]

= \frac{1}{2} \left[ \frac{e}{2} - 1 \right]

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