$\int_{0}^{\pi / 2} \sin 2 x e^{\sin x} d x$ is equal to\

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Professor PC Thomas Classes · Answer

(c) : Let $I=\int_{0}^{\pi / 2} \sin 2 x \cdot e^{\sin x} d x$\ $=\int_{0}^{\pi / 2} 2 \sin x \cos x e^{\sin x} d x \quad(\because \sin 2 x=2 \sin x \cos x)$\ Put $\sin x=t \Rightarrow \cos x d x=d t$\ When $x=0, t=0$; when $x=\pi / 2, t=1$\ $	herefore \quad I=\int_{0}^{1} 2 t e^{t} d t=\left.2 t e^{t}ight|_{0} ^{1}-2 \int_{0}^{1} e^{t} d t$\ $=2 e-\left.2 e^{t}ight|_{0} ^{1}=2 e-2 e+2=2$\