7.1 Integration By Parts/49

Prove $\int _{}^{}\left(\tan ^{n}(x)\right)dx={\frac {\tan ^{n-1}x}{n-1}}-\int _{}^{}\left(\tan ^{n-2}x\right)dx$

$\int _{}^{}\left(\tan ^{n}(x)\right)dx=\int _{}^{}\left((\tan ^{2}x)(\tan ^{n-2}x)\right)dx=\int _{}^{}(\sec ^{2}(x)-1)\tan ^{n-2}(x)dx=\int _{}^{}(\sec ^{2}x)(\tan ^{n-2}x)-\tan ^{n-2}xdx=\int _{}^{}(\sec ^{2}x)(\tan ^{n-2}x)-\int _{}^{}\tan ^{n-2}(x)dx$

Solving for $\int _{}^{}(\sec ^{2}x)(\tan ^{n-2}x)$

${\begin{aligned}&u=\tan ^{n-2}x\quad &dv=\sec ^{2}(x)dx\\[2ex]&du=(n-2)\tan ^{n-3}(x)\cdot \sec ^{2}(x)dx\quad &v=\tan(x)\\[2ex]\end{aligned}}$

${\begin{aligned}\int _{}^{}(\sec ^{2}x)(\tan ^{n-2}x)dx=\tan ^{n-2}(x)\cdot \tan(x)-\int _{}^{}(n-2)\tan ^{n-3}(x)\sec ^{2}\cdot \tan(x)dx\\[2ex]=\tan ^{n-1}(x)-\int _{}^{}(n-2)\tan ^{n-2}(x)\sec ^{2}dx\\[2ex]\tan ^{n-1}(x)-\int _{}^{}(n-2)\tan ^{n-2}(x)\sec ^{2}dx=\int _{}^{}(\sec ^{2}x)(\tan ^{n-2}x)dx\\[2ex]\end{aligned}}$

${\begin{aligned}&&&&&&&&&&&+\int _{}^{}(n-2)\tan ^{n-2}(x)\cdot \sec ^{2}(x)dx\quad &&&&&&+\int _{}^{}(n-2)\tan ^{n-2}(x)\cdot \sec ^{2}(x)dx\end{aligned}}$

</math> \begin{align}

\frac{\tan^{n-1}(x)}{n-1} = \frac{(n-1)}{n-1} \int_{}^{} (\sec^{2}x)(\tan^{n-2}x)dx

\end{align} </math>

Bring down:

</math> \begin{align}

-\int_{}^{}\tan^{n-2}(x)dx

\end{align} </math>

</math> \begin{align}

= \frac{\tan^{n-1}(x)}{n-1} -\int_{}^{}\tan^{n-2}(x)dx

\end{align} </math>

Note: ${\begin{aligned}\tan ^{2}(x)=\sec ^{2}(x)-1\end{aligned}}$