{² x}{( x + x)^{9/2}} \, dx equals

Correct Answer is : −1(sec⁡x+tan⁡x)11/2-{1}{( x + x)¹1/2}}−(secx+tanx)11/21 {111+17(sec⁡x+tan⁡x)}+K≤ft\{ 1/11 + 1/7 ( x + x) \} + K{111+71(secx+tanx)}+K

-{1}{( x + x)¹1/2}} ≤ft\{ 1/11 - 1/7 ( x + x)² \} + K

{1}{( x + x)¹1/2}} ≤ft\{ 1/11 - 1/7 ( x + x)² \} + K

-{1}{( x + x)¹1/2}} ≤ft\{ 1/11 + 1/7 ( x + x) \} + K

{² x}{( x + x)^{9/2}} \, dx equals

Correct Answer is : −1(sec⁡x+tan⁡x)11/2-{1}{( x + x)¹1/2}}−(secx+tanx)11/21 {111+17(sec⁡x+tan⁡x)}+K≤ft\{ 1/11 + 1/7 ( x + x) \} + K{111+71(secx+tanx)}+K

Test Index

Manipal Entrance Test 2015 Math Paper

Question : 26 of 70

Marks: +1, -0

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

equals

$-\frac{1}{(\sec x + \tan x)^{11/2}}$ $\left\{ \frac{1}{11} - \frac{1}{7} (\sec x + \tan x)^2 \right\} + K$
$\frac{1}{(\sec x + \tan x)^{11/2}}$ $\left\{ \frac{1}{11} - \frac{1}{7} (\sec x + \tan x)^2 \right\} + K$
$-\frac{1}{(\sec x + \tan x)^{11/2}}$ $\left\{ \frac{1}{11} + \frac{1}{7} (\sec x + \tan x) \right\} + K$
None of the above

Validate

Solution:

Let

I = \int \frac{\sec^2 x}{(\sec x + \tan x)^{9/2}} \, dx

Put

\sec x + \tan x = t

Then,

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

and

\sec x (\sec x + \tan x) \, dx = dt

\Rightarrow \sec x \, dx = \frac{1}{t} \, dt

Also,

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

\therefore I = \frac{1}{2} \int \frac{ \frac{1}{t} \left( t + \frac{1}{t} \right) }{ t^{9/2} } \, dt

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

\Rightarrow I = -\frac{1}{7 t^{7/2}} - \frac{1}{11 t^{11/2}} + K

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

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

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

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