Question

\[\int \tan^5 x\ dx\]

Answer 1

\[\text{ Let I }= \int \tan^5 \text{ x  dx }\]
\[ = \int \tan^3 x \cdot \tan^2\text{  x dx }\]
\[ = \int \tan^3 x \left( \sec^2 x - 1 \right) dx\]
\[ = \int \tan^3 x \cdot \sec^2 \text{ x dx} - \int \tan^3 \text{ x  dx}\]
\[ = \int \tan^3 x \cdot \sec^2 \text{ x dx} - \int\tan x \cdot \tan^2 \text{ x dx} \]
\[ = \int \tan^3 x \cdot \sec^2 \text{ x dx} - \int\tan x \cdot \left( \sec^2 x - 1 \right) dx\]
\[ = \int \tan^3 x \cdot \sec^2 x dx - \int\tan x \cdot \sec^2\text{ x dx} + \int\tan x dx\]
\[\text{ Putting   tan   x = t   in the  Ist  and IInd integral} . \]
\[ \Rightarrow \sec^2\text{ x dx} = dt\]
\[ \therefore I = \int t^3 \cdot dt - \int t \cdot dt + \int\text{  tan  x  dx }\]
\[ = \frac{t^4}{4} - \frac{t^2}{2} + \text{  ln} \left| \text{ sec  x} \right| + C\]
\[ = \frac{\tan^4 x}{4} - \frac{\tan^2 x}{2} + \text{ ln} \left| \text{  sec  x }\right| + C \left[ \because t = \text{ tan x} \right]\]