Question

Suppose $\int \frac{1-7{\cos }^{2}x}{{\sin }^{7}x{\cos }^{2}x}dx=\frac{g\left(x\right)}{{\sin }^{7}x}+C,$ where $C$ is arbitrary constant of integration. Then the value of $g^{\prime }\left(0\right)+g^{\prime \prime }\left(\frac{\pi }{4}\right)$ is

Answer 1

$\int \frac{1-7{\cos }^{2}x}{{\sin }^{7}x{\cos }^{2}x}dx$
$=\int \frac{{\sec }^{2}x}{{\sin }^{7}x}dx-7\int \frac{1}{{\sin }^{7}x}dx=I_1-I_2$

Now, $I_1=\int \left(\frac{1}{{\sin }^{7}x}\right){\sec }^{2}xdx$
$=\frac{\tan x}{{\sin }^{7}x}+7\int \frac{\tan x}{{\sin }^{8}x}\cos xdx=\frac{\tan x}{{\sin }^{7}x}+I_2$
$\therefore I_1-I_2=\frac{\tan x}{{\sin }^{7}x}+C$, where $C$ is constant of integration.

Hence, $g\left(x\right)=\tan x$
$g^{\prime }\left(x\right)={\sec }^{2}x$ and $g^{\prime \prime }\left(x\right)=2{\sec }^{2}x\tan x$
$\therefore g^{\prime }\left(0\right)=1$ and $g^{\prime \prime }\left(\frac{\pi }{4}\right)=4$
Hence, $g^{\prime }\left(0\right)+g^{\prime \prime }\left(\frac{\pi }{4}\right)=5$