Question

${\int }_0^{\pi }\frac{co{s}^{4}x}{co{s}^{4}x+si{n}^{4}x}dx$ 1s equal to

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

Answer 1

Answer: (B)

Let $I={\int }_0^{\pi }\frac{{\cos }^{4}x}{{\sin }^{4}x+{\cos }^{4}x}dx$
$I=2{\int }_0^{\pi /2}\frac{{\cos }^{4}x}{{\sin }^{4}x+{\cos }^{4}x}dx\dots \text{ (i) }$
$\left[\because {\int }_0^{2a}f(x)dx=2{\int }_0^{a}f(x)dx,\text{ if }f(2a-x)=f(x)\right]$
$\Rightarrow I=2{\int }_0^{\pi /2}\frac{{\cos }^4\left(\frac{\pi }{2}-x\right)}{{\sin }^4\left(\frac{\pi }{2}-x\right)+{\cos }^4\left(\frac{\pi }{2}-x\right)}dx$
$\Rightarrow I=2{\int }_0^{\pi /2}\frac{{\sin }^{4}x}{{\cos }^{4}x+{\sin }^{4}x}dx...(i)$
On adding Eqs. (i) and (ii), we get
$2I=2{\int }_0^{\pi /2}\frac{{\sin }^{4}x+{\cos }^{4}x}{{\sin }^{4}x+{\cos }^{4}x}dx$
$=2{\int }_0^{\pi /2}dx=2[x{]}_0^{\pi r}=2\left(\frac{\pi }{2}-0\right)$
$\Rightarrow 2I=\pi$
$\Rightarrow I=\frac{\pi }{2}$