Mathematics - Indefinite Integration Question with Solution | TestHub

Given,

$I\left(x\right)=\int e^{{\sin }^{2}x}\left(\cos x \sin 2x-\sin x\right)dx$

$\Rightarrow I\left(x\right)=\int e^{{\sin }^{2}x}\left(\cos x-\frac{1}{2 \cos x}\right)\sin 2x dx$

Now let, ${\sin }^{2}x=t\Rightarrow \sin 2x dx=dt$

$I\left(x\right)=\int e^t\left(\sqrt{1-t}-\frac{1}{2\sqrt{1-t}}\right)dt$

Now comparing with,

$I\left(x\right)=\int e^t\left(f\left(t\right)+f^{\text{'}}\left(t\right)\right)dt=e^{t}f\left(t\right)+c$

Here, $f\left(t\right)=\sqrt{1-t} \& f^{\text{'}}\left(t\right)=-\frac{1}{2\sqrt{1-t}}$

$\Rightarrow I\left(x\right)=e^t\sqrt{1-t}+c=e^{{\sin }^{2}x}·\cos x+c$

Now using, $I\left(0\right)=1\Rightarrow c=0$

Hence, the value of $I\left(\frac{\pi }{3}\right)=\frac{1}{2}{e}^{\frac{3}{4}}$