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Evaluate the definite integrals\[\int\limits_0^\frac{\pi}{4}\frac{\sin x\cos x}{\cos^4x+\sin^4x}dx\]

1 Answer

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


answered Feb 19, 2013 by meena.p