Q76 / 170 · Manipal (MET) 2015 · Mathematics

$\int \frac{\sec ^2 x}{(\sec x+\tan x)^{9 / 2}} d x$ equals

  • A
    $-\frac{1}{(\sec x+\tan x)^{11 / 2}}\left\{\frac{1}{11}-\frac{1}{7}(\sec x+\tan x)^2\right\}+K$
  • B
    $\frac{1}{(\sec x+\tan x)^{1 / 12}}\left\{\frac{1}{11}-\frac{1}{7}(\sec x+\tan x)^2\right\}+K$
  • Correct answer
    C
    $-\frac{1}{(\sec x+\tan x)^{11 / 2}}\left\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x)\right\}+K$
  • D
    None of the above

Solution

Let $I=\int \frac{\sec ^2 x}{(\sec x+\tan x)^{9 / 2}} d x$ Put $\sec x+\tan x=t$ Then, $\quad \sec x-\tan x=\frac{1}{t}$ and $\quad \sec x(\sec x+\tan x) d x=d t$ $\Rightarrow \quad \sec x d x=\frac{1}{t} d t$ Also, $\quad \sec x=\frac{1}{2}\left(t+\frac{1}{t}\right)$ $\therefore \quad I=\frac{1}{2} \int \frac{\frac{1}{t}\left(t+\frac{1}{t}\right)}{t^{9 / 2}} d t$ $\begin{aligned} & \Rightarrow \quad I=\frac{1}{2} \int \frac{1}{t^{9 / 2}}+\frac{1}{t^{13 / 2}} d t \\ & \Rightarrow \quad I=-\frac{1}{7 t^{7 / 2}}-\frac{1}{11 t^{11 / 2}}+K \\ & \Rightarrow \quad I=-\frac{1}{t^{11 / 2}\left\{\frac{t^2}{7}+\frac{1}{11}\right\}+K} \\ & \Rightarrow \quad I=-\frac{1}{(\sec x+\tan x)^{11 / 2}} \\ & \left\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x)^2\right\}+K\end{aligned}$

Hence, the correct answer is option: 'C'.

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