Question

$\displaystyle\int_{0}^\pi \frac {cos^4x}{cos^4x+sin^4x}dx$ 1s equal to

• A. $\frac {\pi}{4}$
• B. $\frac {\pi}{2}$
• C. $\frac {\pi}{8}$
• D. $\pi$

Answer 1

Answer: (B)

Let $ I=\displaystyle\int_{0}^{\pi} \frac{\cos ^{4} x}{\sin ^{4} x+\cos ^{4} x} d x $
$ I=2 \displaystyle\int_{0}^{\pi / 2} \frac{\cos ^{4} x}{\sin ^{4} x+\cos ^{4} x} d x\,\,\,\,\,\ldots \text { (i) } $
$\left[\because \displaystyle\int_{0}^{2 a} f(x) d x=2 \int_{0}^{a} f(x) d x, \text { if } f(2 a-x)=f(x)\right] $
$\Rightarrow I=2 \displaystyle\int_{0}^{\pi / 2} \frac{\cos ^{4}\left(\frac{\pi}{2}-x\right)}{\sin ^{4}\left(\frac{\pi}{2}-x\right)+\cos ^{4}\left(\frac{\pi}{2}-x\right)} d x $
$\Rightarrow I=2 \displaystyle\int_{0}^{\pi / 2} \frac{\sin ^{4} x}{\cos ^{4} x+\sin ^{4} x} d x\,\,\,\,\,\,\,\,...(i)$
On adding Eqs. (i) and (ii), we get
$2 I =2 \displaystyle\int_{0}^{\pi / 2} \frac{\sin ^{4} x+\cos ^{4} x}{\sin ^{4} x+\cos ^{4} x} d x $
$=2 \displaystyle\int_{0}^{\pi / 2} d x=2[x]_{0}^{\pi\,r}=2\left(\frac{\pi}{2}-0\right) $
$\Rightarrow 2 I =\pi $
$ \Rightarrow I=\frac{\pi}{2}$