Question

Evaluate the integral
${\int }_0^{2\pi }\frac{1}{1+{\tan }^{4}x}dx$

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

Answer 1

The correct option is D $\pi$

$I={\int }_0^{2\pi }\frac{1}{1+ta{n}^{4}x}dx$

$I=4{\int }_0^{\frac{\pi }{2}}\frac{1}{1+ta{n}^{4}x}dx$

$I_1={\int }_0^{\frac{\pi }{2}}\frac{1}{1+ta{n}^{4}x}dx$ ---(1)

$I_1={\int }_0^{\frac{\pi }{2}}\frac{1}{1+co{t}^{4}x}dx$

$I_1={\int }_0^{\frac{\pi }{2}}\frac{ta{n}^{4}x}{1+ta{n}^{4}x}dx$ ----(2)

$2I_1=\frac{\pi }{2}$

$I_1=\frac{\pi }{4}$

Thus $I=4I_1=\pi$