Question

Suppose $\int \frac{1-7{\cos }^{2}x}{{\sin }^{7}x{\cos }^{2}x}dx=\frac{g(x)}{{\sin }^{7}x}+C$,
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

Answer: 5

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