Question

Evaluate: ${\int }_0^{\pi /4}\frac{\sin x+\cos x}{16+9\sin 2x}dx$.

Answer 1

Let $I={\int }_0^{\frac{\pi }{4}}\frac{\sin x+\cos x}{16+9\sin 2x}dx$

$\sin 2x=2\sin x\cos x$

$\Rightarrow 1-\sin 2x=1-2\sin x\cos x$

$\Rightarrow 1-\sin 2x={\sin }^{2}x+{\cos }^{2}x-2\sin x\cos x$

$\Rightarrow \sin 2x=1-{(\sin x-\cos x)}^2$

$\Rightarrow 9\sin 2x=9-9{(\sin x-\cos x)}^2$

$\Rightarrow 16+9\sin 2x=25-9{(\sin x-\cos x)}^2$

$\therefore I={\int }_0^{\frac{\pi }{4}}\frac{\sin x+\cos x}{25-9{(\sin x-\cos x)}^2}dx$

Let $t=9(\sin x-\cos x)$

$\Rightarrow dt=9(\sin x+\cos x)dx$

When, $x=0$ $\Rightarrow t=-9$

$x=\frac{\pi }{4}\Rightarrow t=0$

$I=\frac{1}{9}{\int }_{-9}^0\frac{dt}{25-t^2}$

$I=\frac{1}{9}{\int }_{-9}^0\frac{dt}{5^2-t^2}$

$I=\frac{1}{90}{\left[\ln \left(\frac{5+t}{5-t}\right)\right]}_{-9}^0$

$I=\frac{1}{90}(\ln 1-\ln \frac{-4}{14})$

$I=-\frac{1}{90}\ln \frac{-2}{7}$