IIT-JEE problem

Level 2

The integral $\int \frac{\sec^2x}{(\sec x+\tan x)^{9/2}}dx$ equals (for some arbitrary constant $K$ )?

-\frac{1}{(\sec x+\tan x)^{11/2}}\left \{\frac{1}{11}-\frac{1}{7}(\sec x+\tan x)^2\right \}+K

\frac{1}{(\sec x+\tan x)^{11/2}}\left \{\frac{1}{11}-\frac{1}{7}(\sec x+\tan x)^2\right \}+K

- \frac{1}{(\sec x+\tan x)^{11/2}}\left \{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x)^2 \right \}+K

\frac{1}{(\sec x+\tan x)^{11/2}}\left \{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x)^2\right \}+K

1 solution

Anish Puthuraya
Feb 4, 2014

Put $\displaystyle \sec x + \tan x = t$

$\Rightarrow\displaystyle (\sec x\tan x + \sec^2x)dx=dt$

$\Rightarrow\displaystyle \sec x dx = \frac{dt}{t}$

Also, note that,
$\displaystyle \sec x-\tan x = \frac{1}{t}$

$\Rightarrow\displaystyle \sec x = \frac{1}{2}(t+\frac{1}{t})$

Hence, the integral simplifies to,

$\displaystyle I = \int \frac{\sec x\sec x dx}{(\sec x+\tan x)^{\frac{9}{2}}} = \int \frac{\frac{1}{2}(t+\frac{1}{t})\frac{dt}{t}}{t^{\frac{9}{2}}}$

This integral is rather simple to evaluate, and it evaluates to,

$\displaystyle I = \frac{-1}{2}(\frac{2}{7t^{\frac{7}{2}}}+\frac{2}{11t^{\frac{11}{2}}})+K$

Back substituting the value of $\displaystyle t = \sec x +\tan x$ ,

$\displaystyle I = -\frac{1}{(\sec x+\tan x)^{\frac{11}{2}}}(\frac{1}{11}+\frac{1}{7}(\sec x+\tan x)^2)+K$