If $\int_{0}^{\pi}\left(\sin ^{3} \mathrm{x}\right) \mathrm{e}^{-\sin ^{2} \mathrm{x}} \mathrm{dx}=\alpha-\frac{\beta}{\mathrm{e}} \int_{0}^{1} \sqrt{\mathrm{t}} \mathrm{e}^{\mathrm{t}} \mathrm{dt}$, then $\alpha+\beta$ is equal to ______

Definite Integrals — JEE Main · 2021 · 27 Jul 2021 · Shift 2 Question with Solution

\int_{0}^{π} (sin^{3} x) e^{- s i n^{2} x} dx = α - \frac{β}{e} \int_{0}^{1} t e^{t} dt

, then

α + β

is equal to ______

Answer: 5

The integrand is unchanged under

x \mapsto π - x

, so

I = \int_{0}^{π} sin^{3} x e^{- s i n^{2} x} d x = 2 \int_{0}^{π /2} sin^{3} x e^{- s i n^{2} x} d x

Writing

sin^{2} x = 1 - cos^{2} x

and substituting

u = cos x

d u = - sin x d x

I = 2 \int_{0}^{π /2} (1 - cos^{2} x) e^{c o s^{2} x - 1} sin x d x = \frac{2}{e} \int_{0}^{1} (1 - u^{2}) e^{u^{2}} d u

Integration by parts gives

\int_{0}^{1} e^{u^{2}} d u = [u e^{u^{2}}]_{0}^{1} - \int_{0}^{1} 2 u^{2} e^{u^{2}} d u = e - 2 \int_{0}^{1} u^{2} e^{u^{2}} d u

and with

t = u^{2}

d u = \frac{d t}{2 t}

\int_{0}^{1} u^{2} e^{u^{2}} d u = \frac{1}{2} \int_{0}^{1} t e^{t} d t

\int_{0}^{1} e^{u^{2}} d u = e - \int_{0}^{1} t e^{t} d t

. Combining,

\int_{0}^{1} (1 - u^{2}) e^{u^{2}} d u = (e - \int_{0}^{1} t e^{t} d t) - \frac{1}{2} \int_{0}^{1} t e^{t} d t = e - \frac{3}{2} \int_{0}^{1} t e^{t} d t

and therefore

I = 2 - \frac{3}{e} \int_{0}^{1} t e^{t} d t

This gives

α = 2

and

β = 3

, so

α + β = 5

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Integer answer