Question

If $U_n={\int }_0^{\pi /4}ta{n}^n\theta d\theta$, then $u_{10}+u_{12}$ is equal to:

• A. $\frac{1}{10}$
• B. $\frac{1}{12}$
• C. $\frac{1}{11}$
• D. $\frac{1}{22}$

Answer 1

The correct option is C $\frac{1}{11}$

$U_n={\int }_0^{\pi /4}{\tan }^n\theta d\theta$

$U_{10}+U_{12}={\int }_0^{\pi /4}{\tan }^{10}\theta d\theta +{\int }_0^{\pi /4}{\tan }^{12}\theta d\theta$

$={\int }_0^{\pi /4}{\tan }^{10}\theta (1+{\tan }^2\theta )d\theta$

$={\int }_0^{\pi /4}{\tan }^{10}\theta {\sec }^2\theta d\theta$

Let, $z=\tan \theta$ $⟹dz={\sec }^2\theta d\theta$

And when $\theta =0,z=0$ and when $\theta =\frac{\pi }{4},z=1$

Now, integration becomes:-

${\int }_0^1{z}^{10}dz=\frac{1}{11}{\left[z^{11}\right]}_0^1$

$=\frac{1}{11}$