The integral ∫sec2⁡xdx(sec⁡x+tan⁡x)92 equals (for some arbitrary constant K)

Let t=sec⁡x+tan⁡x,1t=sec⁡x−tan⁡xt−1t=2tan⁡x⇒121+1t2dt=sec2⁡xdx∴∫sec2⁡xdx(sec⁡x+tan⁡x)92=∫121+1t21t92dt=12∫t−92+t−132dt=12−27t−72−211t−112+K=−t−112111+17t2+K=−1(sec⁡x+tan⁡x)112111+17(sec⁡x+tan⁡x)2+K

The integral ∫sec2⁡xdx(sec⁡x+tan⁡x)92 equals (for some arbitrary constant K)

Let t=sec⁡x+tan⁡x,1t=sec⁡x−tan⁡xt−1t=2tan⁡x⇒121+1t2dt=sec2⁡xdx∴∫sec2⁡xdx(sec⁡x+tan⁡x)92=∫121+1t21t92dt=12∫t−92+t−132dt=12−27t−72−211t−112+K=−t−112111+17t2+K=−1(sec⁡x+tan⁡x)112111+17(sec⁡x+tan⁡x)2+K

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The integral $\int \frac{\sec^{2} x d x}{(\sec x + \tan x)^{\frac{9}{2}}}$ equals (for some arbitrary constant K)

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

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

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

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

answer is C.

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Detailed Solution

Let $t = \sec x + \tan x, \frac{1}{t} = \sec x - \tan x$

$\begin{aligned} t - \frac{1}{t} = 2 \tan x \Rightarrow \frac{1}{2} (1 + \frac{1}{t^{2}}) d t = \sec^{2} x d x \\ ∴ \int \frac{\sec^{2} x d x}{(\sec x + \tan x)^{\frac{9}{2}}} = \int \frac{1}{2} (1 + \frac{1}{t^{2}}) \frac{1}{t^{\frac{9}{2}}} d t \end{aligned}$

$\begin{array}{l} = \frac{1}{2} \int (t^{- \frac{9}{2}} + t^{- \frac{13}{2}}) d t = \frac{1}{2} [- \frac{2}{7} t^{- \frac{7}{2}} - \frac{2}{11} t^{- \frac{11}{2}}] + K \\ = - t^{- \frac{11}{2}} [\frac{1}{11} + \frac{1}{7} t^{2}] + K \\ = - \frac{1}{(\sec x + \tan x)^{\frac{11}{2}}} \{\frac{1}{11} + \frac{1}{7} (\sec x + \tan x)^{2}\} + K \end{array}$