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 a constant of integration. Then the value of $g^{\prime }\left(0\right)+g^{\prime \prime }\left(\frac{\pi }{4}\right)$ is

• A. $0$
• B. $1$
• C. $3$
• D. $5$

Answer 1

The correct option is D $5$

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

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

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

Alternatively :
We have $\int \frac{1-7{\cos }^{2}x}{{\sin }^{7}x{\cos }^{2}x}dx=\frac{g\left(x\right)}{{\sin }^{7}x}+C$
Differentiating both sides with respect to $x$, we get
$\frac{1-7{\cos }^{2}x}{{\sin }^{7}x{\cos }^{2}x}=\frac{\left({\sin }^{7}x\right)g^{\prime }\left(x\right)-g\left(x\right)\left(7{\sin }^{6}x\cos x\right)}{{\sin }^{14}x}$
$\Rightarrow {\sec }^{2}x-7=g^{\prime }\left(x\right)-7g\left(x\right)\cot x$, which is possible when $g\left(x\right)=\tan x$

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