Question

If $\int \frac{dx}{\sin x\cdot \cos x({\tan }^{9}x+1)}=\frac{1}{k}\ln \left|\frac{(\sin x{)}^9}{(\sin x{)}^9+(\cos x{)}^9}\right|+C$, then the value of $k^2+1$ is:
(where $C$ is integration constant)

Answer 1

$I=\int \frac{dx}{\sin x\cdot \cos x({\tan }^{9}x+1)}=\int \frac{dx}{\tan x\cdot {\cos }^{2}x({\tan }^{9}x+1)}=\int \frac{{\sec }^{2}xdx}{\tan x({\tan }^{9}x+1)}$
Put $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$
$\Rightarrow I=\int \frac{dt}{t(t^9+1)}=\int \frac{dt}{t^{10}(1+t^{-9})}$
Taking $1+t^{-9}=y\Rightarrow -9t^{-10}dt=dy$
$\Rightarrow I=\int \frac{dy}{-9y}=-\frac{1}{9}\ln |y|+C\Rightarrow I=-\frac{1}{9}\ln |1+t^{-9}|+C\Rightarrow I=-\frac{1}{9}\ln \left|\frac{(\tan x{)}^9+1}{(\tan x{)}^9}\right|+C$
$\Rightarrow I=\frac{1}{9}\ln \left|\frac{(\sin x{)}^9}{(\sin x{)}^9+(\cos x{)}^9}\right|+C$
Hence, $k=9\Rightarrow k^2+1=82$