Question

If $x\in \left(\frac{\pi }{4},\frac{3\pi }{4}\right),$ then $\int \frac{\sin x-\cos x}{\sqrt{1-\sin 2x}}{e}^{\sin x}$ $\cos xdx=$

• A. $e^{\sin x}+c$
• B. $e^{\sin x-\cos x}+c$
• C. $e^{\sin x+\cos x}+c$
• D. $e^{-\sin x-\cos x}+c$

Answer 1

The correct option is A $e^{\sin x}+c$

Solution :

Given $x\in (\frac{\pi }{4},\frac{3\pi }{4})$

$I=\int \frac{\sin x-\cos x}{\sqrt{1-\sin 2x}}{e}^{\sin x}\cos xdx$

we have $\sqrt{1-\sin 2x}=\sqrt{{\sin }^{2}x+{\cos }^{2}x-2\sin x\cos x}$

$\sqrt{(\sin x-\cos x{)}^2}=\sin x-\cos x$

$\therefore I=\int e^{\sin x}\cos xdx$

Now let $\sin x-1$

$\Rightarrow \cos dx=dt$

$\therefore I=\int e^{t}dt$

$=e^t+C$

Substituting value of $t,$ we get

$I=e^{\sin x}+C$