Question

The value of $\underset{1}{\overset{2}{\int }}(3x^2\tan \frac{1}{x}-x{\sec }^2\frac{1}{x})dx$ is

• A. $8\tan \frac{1}{2}$
• B. $\tan 1$
• C. $8\tan \frac{1}{2}-\tan 1$
• D. $8\tan \frac{1}{2}-3\tan 1$

Answer 1

The correct option is C $8\tan \frac{1}{2}-\tan 1$

Let $I=\underset{1}{\overset{2}{\int }}(3x^2\tan \frac{1}{x}-x{\sec }^2\frac{1}{x})dx$
Using ILATE in first integration, treating $\tan \left(\frac{1}{x}\right)$ as first function, we get
$I={x^3\tan \left(\frac{1}{x}\right)|}_1^2+\underset{1}{\overset{2}{\int }}x{\sec }^2\left(\frac{1}{x}\right)dx-\underset{1}{\overset{2}{\int }}x{\sec }^2\left(\frac{1}{x}\right)dx={x^3\tan \frac{1}{x}|}_1^2$
$\therefore I=8\tan \frac{1}{2}-\tan 1$