Question

If $\int \frac{1-7{\cos }^{2}x}{{\sin }^{7}x\cdot {\cos }^{2}x}dx=\frac{g^{(x)}}{{\sin }^{7}x}+c$, then $g^{\prime }(0)+g^{\prime \prime }\left(\frac{\pi }{4}\right)$ is

• A. $1$
• B. $17$
• C. $9$
• D. $5$

Answer 1

Answer: (D)

$\begin{array}{l}\int \frac{1-7{\cos }^{2}x}{{\sin }^{7}x\cdot {\cos }^{2}x}dx=\frac{g(x)}{{\sin }^{7}x}+c,\\ =\int \frac{1}{{\sin }^{7}x{\cos }^{2}x}dx-7\int \frac{{\cos }^{2}x}{{\sin }^{7}x{\cos }^{2}x}dx\\ =\int {\mathrm{cosec}}^{7}x{\sec }^{2}xdx-7\int {\mathrm{cosec}}^{7}xdx\end{array}$
Assume, $I=\int {\mathrm{cosec}}^{7}x{\sec }^{2}xdx$
By integration by parts
$\begin{array}{l}I={\mathrm{cosec}}^{7}x\tan x-\int 7{\mathrm{cosec}}^{7}xdx\\ =\frac{\tan x}{{\sin }^{7}x}+7\int \frac{1}{{\sin }^{7}x}+c\end{array}$
So,
$\int {\mathrm{cosec}}^{7}x{\sec }^{2}xdx-7\int {\mathrm{cosec}}^{7}xdx={\mathrm{cosec}}^{7}x\tan x+\int 7{\mathrm{cosec}}^{7}xdx-\int 7{\mathrm{cosec}}^{7}xdx$
Hence,
$g(x)=\tan x$
$g^{\prime }(x)={\sec }^{2}x$
$g^{\prime \prime }(x)=2{\sec }^{2}x\tan x$
$g^{\prime }(0)+g^{\prime \prime }\left(\frac{\pi }{4}\right)=5$