Question

Write a value of\[\int\frac{\sec^2 x}{\left( 5 + \tan x \right)^4} dx\]

Answer 1

\[\text{ Let I }= \int\frac{\sec^2 x dx}{\left( 5 + \tan x \right)^4}\]
\[\text{ Let 5 + tan  x = t }\]
\[ \Rightarrow \sec^2 x \text{ dx } = dt\]
\[ \therefore I = \int\frac{dt}{t^4}\]
\[ = \int t^{- 4} dt\]
\[ = \left[ \frac{t^{- 4 + 1}}{- 4 + 1} \right] + C\]
\[ = - \frac{1}{3 t^3} + C\]
\[ = - \frac{1}{3 \left( 5t + \tan x \right)^3} + C \left( \because t = 5 + \tan x \right)\]