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Q.
If $\int \frac{1-7 \cos ^2 x}{\sin ^7 x \cdot \cos ^2 x} d x=\frac{g^{(x)}}{\sin ^7 x}+c$, then $g^{\prime}(0)+g^{\prime \prime}\left(\frac{\pi}{4}\right)$ is
NTA AbhyasNTA Abhyas 2022
Solution:
$
\begin{array}{l}
\int \frac{1-7 \cos ^2 x}{\sin ^7 x \cdot \cos ^2 x} d x=\frac{g(x)}{\sin ^7 x}+c, \\
=\int \frac{1}{\sin ^7 x \cos ^2 x} d x-7 \int \frac{\cos ^2 x}{\sin ^7 x \cos ^2 x} d x \\
=\int \operatorname{cosec}^7 x \sec ^2 x d x-7 \int \operatorname{cosec}^7 x d x
\end{array}
$
Assume, $I=\int \operatorname{cosec}^7 x \sec ^2 x d x$
By integration by parts
$
\begin{array}{l}
I=\operatorname{cosec}^7 x \tan x-\int 7 \operatorname{cosec}^7 x d x \\
=\frac{\tan x}{\sin ^7 x}+7 \int \frac{1}{\sin ^7 x}+c
\end{array}
$
So,
$
\int \operatorname{cosec}^7 x \sec ^2 x d x-7 \int \operatorname{cosec}^7 x d x=\operatorname{cosec}^7 x \tan x+\int 7 \operatorname{cosec}^7 x d x-\int 7 \operatorname{cosec}^7 x d x
$
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
$