Question

If the reduction formula for $I_n=\int {\tan }^n\mathrm{xdx}$ is given by
$I_n=\frac{1}{n-1}{\tan }^{n-1}x-I_{n-2}$, then $\int {\tan }^{3}x\mathrm{dx}\mathrm{is}$

• A.
  $\frac{1}{2}{\tan }^{2}x-\frac{1}{2}\ln \left|1+{\tan }^{2}x\right|+C$
• B.
  $-\frac{1}{2}{\tan }^{2}x-\frac{1}{2}\ln \left|1+{\tan }^{2}x\right|+C$
• C.
  $\frac{1}{2}{\tan }^{2}x+\frac{1}{2}\ln \left|1+{\tan }^{2}x\right|+C$
• D.
  $-\frac{1}{2}{\tan }^{2}x+\frac{1}{2}\ln \left|1+{\tan }^{2}x\right|+C$

Answer 1

The correct option is A

$\frac{1}{2}{\tan }^{2}x-\frac{1}{2}\ln \left|1+{\tan }^{2}x\right|+C$
We know the reduction formulae for $I_n=\int {\tan }^{n}x\mathrm{dx}$ as:
$I_n=\frac{1}{n-1}{\tan }^{n-1}x-I_{n-2}$
Now, to find $\int {\tan }^{3}x\mathrm{dx}$
we assume $n=3$.
Then, we can write the reduction formulae as:
$I_3=\frac{1}{3-1}{\tan }^{3-1}x-I_{3-2}$
$\Rightarrow I_3=\frac{1}{2}{\tan }^{2}x-\int \mathrm{tanx}\mathrm{dx}$
Now, we know
$\int \tan x\mathrm{dx}=\ln \left(\mathrm{secx}\right|+C\Rightarrow \int \mathrm{tanx}\mathrm{dx}=\ln \left(\sqrt{1+{\tan }^{2}x}\right|+C\Rightarrow I_1=\frac{1}{2}\ln \left(1+{\tan }^{2}x\right|+C$
Thus, substituting for $I_1$ we get,

$\int {\tan }^{3}x\mathrm{dx}=\frac{1}{2}{\tan }^{2}x-\frac{1}{2}\ln \left(1+{\tan }^{2}x\right|+C$
Thus, The option a. is correct.