If 2 sec x tan x ( sec x - tan x ) 10 d x = ( sec x + tan x ) 11 [ p ( sec x - tan x ) 2 + q ] + c , then ( 1 p + 1 q ) is equal to (where ' c ' is constant of integration)

Question

If $\int \frac{2 \sec x \tan x}{{(\sec x - \tan x)}^{10}} d x = {(\sec x + \tan x)}^{11} [p {(\sec x - \tan x)}^{2} + q] + c$ , then $(\frac{1}{p} + \frac{1}{q})$ is equal to (where $' c'$ is constant of integration)

Smart Achievers · Accepted Answer

(3)

$Let \sec x - \tan x = t$

$then (2 \sec x \tan x) d x = (1 - \frac{1}{t^{2}}) d t$

$I = - \frac{1}{9 t^{9}} + \frac{1}{11 t^{11}} + C$

$\Rightarrow \frac{1}{p} + \frac{1}{q} = 2$

Solution

Q.
If $\int \frac{2 \sec x \tan x}{{(\sec x - \tan x)}^{10}} d x = {(\sec x + \tan x)}^{11} [p {(\sec x - \tan x)}^{2} + q] + c$ , then $(\frac{1}{p} + \frac{1}{q})$ is equal to (where $' c'$ is constant of integration)

1 1

2 - 1

3 2

4 - 2

Ans.
(3)

$Let \sec x - \tan x = t$

$then (2 \sec x \tan x) d x = (1 - \frac{1}{t^{2}}) d t$

$I = - \frac{1}{9 t^{9}} + \frac{1}{11 t^{11}} + C$

$\Rightarrow \frac{1}{p} + \frac{1}{q} = 2$

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