Mathematics

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


ANSWER

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


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
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