Question

Evaluate: $\int \frac{\sin x+\cos x}{\sqrt{9+16\sin 2x}}dx$

Answer 1

Here, $\int \frac{\sin x+\cos x}{\sqrt{9+16\sin 2x}}dx$
$\int \frac{\sin x+\cos x}{\sqrt{9+16\sin 2x+16-16}}dx=\int \frac{\sin x+\cos x}{\sqrt{25-16(1-\sin 2x)}}dx$
$=\int \frac{\sin x+\cos x}{\sqrt{25-16({\sin }^{2}x+{\cos }^{2}x-\sin 2x)}}dx$
$=\int \frac{\sin x+\cos x}{\sqrt{25-16{(\sin x-\cos x)}^2}}dx$
Let $\sin x-\cos x=t$
$(\cos x+\sin x)dx=dt$
$\int \frac{dt}{\sqrt{{\left(5\right)}^2-{\left(4t\right)}^2}}=\frac{1}{4}{\sin }^{-1}\left(\frac{4t}{5}\right)+c$
[$\because \int \frac{1}{\sqrt{a^2-x^2}}dx={\sin }^{-1}\left(\frac{x}{a}\right)+c$]
Putting the value of $t$, we get
$=\frac{1}{4}{\sin }^{-1}\left[\frac{4(\sin x-\cos x)}{5}\right]+c$