Question

${\int }_{-\pi }^{\pi }\frac{{\sin }^{4}x}{{\sin }^{4}x+{\cos }^{4}x}dx$ is equal to:

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{2}$
• C. $\frac{3\pi }{2}$
• D. $\frac{\pi }{2}$
• E. $\frac{2\pi }{3}$

Answer 1

Answer: (A)

Let $I={\int }_{-\pi }^{\pi }\frac{{\sin }^{4}xdx}{{\sin }^{4}x+{\cos }^{4}x}$
$=4{\int }_0^{\pi }\frac{{\sin }^{4}x}{{\sin }^{4}x+{\cos }^{4}x}dx$
$I=4{\int }_0^{\pi /2}\frac{{\sin }^{4}x}{{\sin }^{4}x+{\cos }^{4}x}dx$ ..(i)
$I=4{\int }_0^{\pi /2}\frac{{\cos }^{4}x}{{\sin }^{4}x+{\cos }^{4}x}dx$ ..(ii)
On adding Eqs. (i) and (ii), we get
$2I=4{\int }_0^{\pi /2}1.dx=2\pi$