The integral ^2 x ( x+ x)^ 9 / 2 d x equals (for some arbitrary c… - Vidyadip

MCQ

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

A
$\frac{-1}{(\sec x+\tan x)^{1 / 2}}\left\{\frac{1}{11}-\frac{1}{7}(\sec x+\tan x)^2\right\}+K$
B
$\frac{1}{(\sec x+\tan x)^{1 / 12}}\left\{\frac{1}{11}-\frac{1}{7}(\sec x+\tan x)^2\right\}+K$
✓
$\frac{-1}{(\sec x+\tan x)^{11 / 2}}\left\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x)^2\right\}+K$
D
$\frac{1}{(\sec x+\tan x)^{11 / 2}}\left\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x)^2\right\}+K$

✓

Answer

Correct option: C.

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

c
$\text { Put } \quad \sec x+\tan x=t $

$\left(\sec x \tan x+\sec ^2 x\right) d x=d t $

$\text { sec } x . tdx = dt $

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

$\sec x=\frac{t+\frac{1}{t}}{2} $

$\int \frac{\sec x . d t}{t^{9 / 2} . t}=\int \frac{1}{2} \frac{\left(t+\frac{1}{t}\right)}{t \cdot t^{9 / 2}} d t $

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

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

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

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