# Evaluate : $\int_0^{\Large\frac{\pi}{4}} \large\frac{\sin\: x.\cos\: x}{\cos^4x+sin^4x}$$dx ## 1 Answer Toolbox: • f(x) is an integral function and we substitute f(x) for t, then f'(x)dx=dt, hence \int f(x)dx=\int t.dt • \int \large\frac{dx}{x^2+a^2}=\large\frac{1}{a}$$\tan ^{-1}(x/a)+c$
Step 1:
$\int\limits_0^{\Large\frac{\pi}{4}}\large\frac{\sin x\cos x}{\cos^4x+\sin^4x}$$dx Multiply and divide by \cos ^4 x I=\Large\int\limits_0^\frac{\pi}{4}\Large\frac{\frac{\sin x\cos x}{\cos^4x}}{\frac{\cos ^4x+\sin^4x}{cos ^4x}} But we know \large\frac{\sin x}{\cos x}$$=\tan x\; and \;\large\frac{1}{\cos x}$$=\sec x Hence \large\int\limits_0^{\Large\frac{\pi}{4}}\large\frac{\tan x\sec ^2x}{1+\tan ^4x}$$dx$
Step 2:
Let $\tan ^2 x=t$
on differentiating w.r.t x ,
$2\tan x.\sec^2x dx=dt$
$\Rightarrow \tan x \sec^2xdx=dt/2$
This limits also change when we substitute t,
When $x=0,t=\tan^2 0=0$
When $x=\large\frac{\pi}{4}$$,t=\tan ^2\large\frac{\pi}{4}$$=1$
Therefore $I=\frac{1}{2}\int _0^1 \large\frac{dt}{1+t^2}$
Step 3:
This is of the form $\int\large \frac{dx}{x^2+a^2}$$=\large\frac{1}{a}$$\tan ^{-1}(x/a)+c$
Therefore $\frac{1}{2}\int_0^1 \large\frac{dt}{1+t^2}=\frac{1}{2}$$[\tan ^{-1}(t)]_0^1+c On applying limits, \Rightarrow \large\frac{1}{2}$$[\tan^{-1}-\tan ^{-1}(0)]$
But $\tan^{-1}=\large\frac{\pi}{4}$
$\qquad\qquad=\large\frac{1}{2} \times \frac{\pi}{4}=\frac{\pi}{8}$
Hence $I=\large\frac{\pi}{8}$