Mathematics

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

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

SOLUTION
$\displaystyle I=\int \dfrac {\sec ^2x}{(\sec x+\tan x)^{\frac {9}{2}}}dx$

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

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

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

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

$\displaystyle \sec x dx=\dfrac {dt}{t}, \dfrac {1}{2}\left(t+\dfrac {1}{t}\right)=\sec x$

$\displaystyle I=\dfrac {1}{2}\int \dfrac {\left(t+\dfrac {1}{t}\right)}{t^{\frac {9}{2}}}\dfrac {dt}{t}$

$\displaystyle =\dfrac {1}{2}\int (t^{\frac {-9}{2}}+t^{\frac {-13}{2}})dt$

$\displaystyle =\dfrac {1}{2}\left[\dfrac {t^{\frac {-9}{2}+1}}{\dfrac {-9}{2}+1}+\dfrac {t^{\frac {-13}{2}+1}}{\dfrac {-13}{2}+1}\right]$

$=-\dfrac {1}{7} \dfrac {1}{t^{\frac {7}{2}}}-\dfrac {1}{11} \dfrac {1}{t^{\frac {11}{2}}}$

$=\dfrac {-1}{t^{\frac {11}{2}}}\left(\dfrac {1}{11}+\dfrac {t^2}{7}\right)$

$=-\dfrac {1}{(\sec x+\tan x)^{\frac {11}{2}}}\left \{\dfrac {1}{11}+\dfrac {1}{7}(\sec x+ \tan x)^2\right \}+k$

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Single Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

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