Question

If $\int {\tan }^{7}xdx=f\left(x\right)+\log \left|\cos x\right|$, then $f\left(x\right)$ is polynomial in $\tan x$ of degree.

Answer 1

$\int ta{n}^{7}xdx$

put $tanx=t$

$dx=\frac{dt}{1+t^2}$

$=\int \frac{t^{7}dt}{1+t^2}=\int \left[t^5-t^3+t-\frac{t}{t^2+1}\right]dt=\int \left(t^5-t^3+t\right)dt-\frac{1}{2}\int \frac{2t}{t^2+1}dt$

$=\frac{t^6}{6}-\frac{t^4}{4}+\frac{t^2}{2}-\frac{1}{2}\log \left|t^2+1\right|+C$

$=\frac{{\tan }^{6}x}{6}-\frac{{\tan }^{4}x}{4}+\frac{{\tan }^{2}x}{2}-\frac{1}{2}\log \left|{\tan }^{2}x+1\right|+C$

$=\frac{{\tan }^{6}x}{6}-\frac{{\tan }^{4}x}{4}+\frac{{\tan }^{2}x}{2}-\frac{1}{2}\times 2\log \left|\sec x\right|+C$

$=\frac{{\tan }^{6}x}{6}-\frac{{\tan }^{4}x}{4}+\frac{{\tan }^{2}x}{2}-\log \left|\sec x\right|+C$

$=f\left(x\right)+\log |cosx|+C$

where $f\left(x\right)$ is a polynomial in $\tan x$

Therefore,

degree of $f\left(x\right)$ is $6$.