Discussion Forum

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

Answer:

Explanation:

Concept involved Integration by Substitution

i.e,    $I=\int f(g(x)).g'(x) dx$

 put    , g(x)=t $\Rightarrow$ g'(x)dx=dt$

 $\therefore$  $l=\int f(t) dt$

 Description of Situation Generally,

students gets confused after substitution i.e.,

 $\sec x+\tan x=t$

 Now, for sec x  we should use

 $\sec^{2}x-tan^{2}x=1$

 $\Rightarrow$   $(\sec x-\tan x)(\sec x+\tan x)=1$

 $\Rightarrow$  $\sec x-\tan x=\frac{1}{t}$

 sol. Here, $\int  \frac{\sec^{2}x}{(\sec x+\tan x)^{9/2}}dx$

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

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

$\Rightarrow$   $\sec x .t dx=dt$

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

 $\therefore$  $\sec x-\tan x= \frac{1}{t}$

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

 $\therefore$   $l=\int \frac{\sec x.\sec x dx}{(\sec x+\tan x)^{9/2}}$

 $\Rightarrow$   $l=\int \frac{\frac{1}{2}\left(t+\frac{1}{t}\right).\frac{dt}{t}}{t^{9/2}}$

$=\frac{1}{2}\int\left(\frac{1}{t^{9/2}}+\frac{1}{t^{13/2}}\right)dt$

 =$-\frac{1}{2}\left\{\frac{2}{7 t^{7/2}}+\frac{2}{11t^{11/2}}\right\}+K$

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

$=\frac{-1}{(\sec x+\tan x)^{11/2}}$

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