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Evaluate the integral of $tan^{6}x$
Option: 1
$\frac{tan^{5}x}{5}-\frac{tan^{3}}{3}x+tan\ x-x$

Option: 2
$\frac{tan^{5}x}{5}-\frac{tan^{3}}{3}x-tan\ x-x$

Option: 3
$-\frac{tan^{5}x}{5}-\frac{tan^{3}}{3}x-tan\ x-x$

Option: 4
$\frac{tan^{5}x}{5}+\frac{tan^{3}}{3}x+tan\ x+x$

Answers (1)

Given integral,
$\int tan^{6}x\ dx$
Consider,
$I_{n}=\int tan^{n}x\ dx$
Using the reduction method,
Put $n=6,$
$I_{6}=\int tan^{6}dx=\frac{tan^{5}x}{5}-I_{4}..........(1)$
Put $n=4,$
$I_{4}=\int tan^{4}dx=\frac{tan^{3}x}{3}-I_{2}.........(2)$
Put $n=2,$
$I_{2}=\int tan^{2}dx=\frac{tan\ x}{3}-I_{0}..........(3)$
Put $n=0,$
$I_{0}=\int tan^{0}dx=\int dx=x$
Substitute $x=I_{0}$ in equation $\left ( 3 \right )$ ,
$I_{2}=tan\ x-x$
Substitute the value of $I_{2}$ in equation $\left ( 2 \right )$ ,
$I_{4}=\frac{tan^{3}x}{3}-tan\ x+x$
Substitute the value of $I_{4}$ in equation $\left ( 1 \right )$ ,
$I_{6}=\frac{tan^{5}x}{5}-\frac{tan^{3}}{3}x+tan\ x-x$
Therefore,