# Evaluate the definite integrals$\int\limits_0^\frac{\pi}{4}\frac{\sin x+\cos x}{9+16\sin 2x}\;dx$

Toolbox:
• (i)Let $f(x)=t,\;then \;f'(x)=dt \;Therefore\; \int f(x)dx=\int t.dt$
• (ii)$\sin 2x=2\sin x\cos x$
• (iii)$\int \frac{dx}{a^2-x^2}=\frac{1}{2a} log \frac{|a-x|}{|a+x|}+c$
Let I=$\int\limits_0^\frac{\pi}{4}\frac{\sin x+\cos x}{9+16\sin 2x}\;dx$

$=\int\limits_0^\frac{\pi}{4}\frac{\sin x+\cos x}{9+16(2\sin x \cos x)}\;dx$

Let $(\sin x-\cos x)=t$ on differentiating w.r.t x we get , $(\cos x+\sin x)dx=dt$

When we substitute x for t then the limit also changes

When $x=0,t=\sin 0-\cos 0=-1 \qquad(\cos 0=1)$

When $x=\frac{\pi}{4},t=\sin \frac{\pi}{4}-\cos \frac{\pi}{4}=0 \qquad(\sin \frac{\pi}{4}=\cos \frac{\pi}{4}=\frac{1}{\sqrt 2})$

$(\sin x-\cos x)^2=t^2$

On expanding we get,

$\sin^2x+\cos^2x-2\sin x \cos x=t^2$

$1-2\sin x \cos x =t^2$

$=>1-\sin 2x=t^2$

Therefore $\sin 2x=1-t^2$

Now substituting t and dt we get,

$I=\int \limits_{-1}^0 \frac{dt}{9+16(1-t^2)}$