Question

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

Answer 1

We have $\int \frac{1-7 \cos ^2 x}{\sin ^7 x \cos ^2 x} d x=\int \frac{\sec ^2 x}{\sin ^7 x} d x-7 \int \frac{1}{\sin ^7 x} d x=I_1-I_2$
Now, $I_1=\int\left(\frac{1}{\sin ^7 x}\right) \sec ^2 x d x=\frac{\tan x}{\sin ^7 x}+7 \int \frac{\tan x}{\sin ^8 x} \cos x d x=\frac{\tan x}{\sin ^7 x}+I_2$
(I) (II)
(By parts)
$\therefore I_1-I_2=\frac{\tan x}{\sin ^7 x}+C$, where $C$ is constant of integration.
Hence $g(x)=\tan x$
So, $ g^{\prime}(x)=\sec ^2 x$ and $g^{\prime \prime}(x)=2 \sec ^2 x \tan x$
$\therefore g ^{\prime}(0)=1$ and $g ^{\prime \prime}\left(\frac{\pi}{4}\right)=4 . $
Hence $g ^{\prime}(0)+ g ^{\prime \prime}\left(\frac{\pi}{4}\right)=1+4=5$
Alternatively: We have $\int \frac{1-7 \cos ^2 x}{\sin ^7 x \cos ^2 x} d x=\frac{g(x)}{\sin ^7 x}+C$.....(1)
Differentiating both sides with respect to $x$, we get, $\frac{1-7 \cos ^2 x}{\sin ^7 x \cos ^2 x}=\frac{\left(\sin ^7 x\right) g^{\prime}(x)-g(x)\left(7 \sin ^6 x \cos x\right)}{\sin ^{14} x}$ $\Rightarrow \sec ^2 x-7=g^{\prime}(x)-7 g(x) \cot x$, which is possible when $g(x)=\tan x$.
So, $ g^{\prime}(x)=\sec ^2 x$ and $g^{\prime \prime}(x)=2 \sec ^2 x \tan x$
$\therefore g ^{\prime}(0)=1 \text { and } g ^{\prime \prime}\left(\frac{\pi}{4}\right)=4 \text {. Hence } g ^{\prime}(0)+ g ^{\prime \prime}\left(\frac{\pi}{4}\right)=1+4=5 $