Question

Evaluate: $\underset{0}{\overset{\pi /2}{\int }}\frac{\left(\sin x+\cos x\right)dx}{9+16\sin 2x}$

Answer 1

$I=\int \frac{\sin x+\cos x}{9+16\sin 2x}dx$

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

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

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

Now multiplying by $16$

$16\sin 2x=16-16(\sin x-\cos x{)}^2$

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

$\therefore I=\int \frac{sinx+\cos x}{25-16(\sin x-\cos x{)}^2}dx$

Let $4(\sin x-\cos x)=t$

$\Rightarrow 4(\cos x+\sin x)dx=dt$

$\therefore I=\frac{1}{4}\int \frac{dt}{25-t^2}$

$=\frac{1}{4}\int \frac{dt}{5^2-t^2}$

$=\frac{1}{40}log\left|\frac{5+t}{5-t}\right|+c$

$=\frac{1}{40}log\left|\frac{5+4(\sin x-\cos x)}{5-4(\sin x-\cos x)}\right|+c$