The integral ∫sec2⁡x(sec⁡x+tan⁡x)9/2dx equals (for some arbitrary constant K)

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

1. A

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

2. B

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

3. C

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

4. D

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

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### Solution:

We have,

$I=\int \frac{{\mathrm{sec}}^{2}x}{\left(\mathrm{sec}x+\mathrm{tan}x{\right)}^{9/2}}dx$

Let $secx+\mathrm{tan}x=t$.

. Then, $\mathrm{sec}x-\mathrm{tan}x=1/t$ and,

$\mathrm{sec}x\left(\mathrm{sec}x+\mathrm{tan}x\right)dx=dt$

and, $\mathrm{sec}x=\frac{1}{2}\left(t+\frac{1}{t}\right)$

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