Question

If $f\left(x\right)=\cos \left({\tan }^{-1}x\right)$, then the value of the integral ${\int }_0^{1}x{f}^{\prime \prime }\left(x\right)dx$ is

• A. $\frac{3-\sqrt{2}}{2}$
• B. $\frac{3+\sqrt{2}}{2}$
• C. $-1$
• D. $1-\frac{3}{2\sqrt{2}}$

Answer 1

The correct option is C $-1$

$\begin{array}{c}f\left(x\right)=\cos \left({\tan }^{-1}x\right)\to \left(i\right)\\ \Rightarrow {\int }_0^{1}x{f}^{\prime \prime }\left(x\right)dx\to \left(ii\right)\\ put{\tan }^{-1}x=t\\ x=\tan t\end{array}$

Diff. w.r.t. t, we get

$\begin{array}{c}dx={\sec }^{2}tdt\\ f\left(x\right)=\cos t\\ f^{\prime }\left(x\right)=-\sin t\end{array}$

again diff. w.r.t. t,

$f^{\prime \prime }\left(x\right)=-\cos t$

put the value in equation (2), we get

$\begin{array}{c}{\int }_0^1\tan t.\left(-\cos t\right)dt\left({\sec }^{2}t\right)\\ -{\int }_0^1\frac{\sin t}{\cos t}\times \cos t\times {\sec }^{2}tdt\\ {\int }_0^1\frac{\sin t}{{\cos }^{2}t}dt\end{array}$ Now, put $\cos t=\theta -\sin tdt=d\theta$

$\begin{array}{c}-{\int }_0^1-\frac{d\theta }{{\theta }^2}\\ \Rightarrow {\int }_0^1\frac{1}{{\theta }^2}d\theta \\ \Rightarrow {\left[\frac{-1}{\theta }\right]}_0^1\\ \Rightarrow \left[\frac{-1}{1}+\left(\frac{-1}{0}\right)\right]\\ \Rightarrow -1.\end{array}$