Question

$\int \left\{\frac{se{c}^{2}x}{{\left(secx+\tan x\right)}^{{\displaystyle \frac{9}{2}}}}\right\}dx=$ (for some arbitrary constant $c$)

• A. $\frac{-1}{{\left(secx+\tan x\right)}^{{\displaystyle \frac{11}{2}}}}\left\{\frac{1}{11}-\frac{1}{7}{\left(secx+\tan x\right)}^2\right\}+c$
• B. $\frac{1}{{\left(secx+\tan x\right)}^{{\displaystyle \frac{11}{2}}}}\left\{\frac{1}{11}-\frac{1}{7}{\left(secx+\tan x\right)}^2\right\}+c$
• C. $\frac{-1}{{\left(secx+\tan x\right)}^{{\displaystyle \frac{11}{2}}}}\left\{\frac{1}{11}+\frac{1}{7}{\left(secx+\tan x\right)}^2\right\}+c$
• D. $\frac{1}{{\left(secx+\tan x\right)}^{{\displaystyle \frac{11}{2}}}}\left\{\frac{1}{11}+\frac{1}{7}{\left(secx+\tan x\right)}^2\right\}+c$

Answer 1

The correct option is C

$\frac{-1}{{\left(secx+\tan x\right)}^{{\displaystyle \frac{11}{2}}}}\left\{\frac{1}{11}+\frac{1}{7}{\left(secx+\tan x\right)}^2\right\}+c$

Explanation for correct option:

Evaluate $\int \left\{\frac{se{c}^{2}x}{{\left(secx+\tan x\right)}^{{\displaystyle \frac{9}{2}}}}\right\}dx$

Consider the given Equation as

$I=\int \left\{\frac{se{c}^{2}x}{{\left(secx+\tan x\right)}^{{\displaystyle \frac{9}{2}}}}\right\}dx$

Let us assume that,

$secx+\tan x=t......\left(1\right)$

We know that,

$$\begin{gathered} se{c}^{2}x-{\tan }^{2}x=1 \\ \Rightarrow \left(secx-\tan x\right)\left(secx+\tan x\right)=1 \\ \Rightarrow \left(secx-\tan x\right)=\frac{1}{t}......\left(2\right) \end{gathered}$$

Add the above Equations

$$\begin{gathered} secx+\tan x=t \\ secx-\tan x=t \\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ 2secx=t+\frac{1}{t} \\ secx=\frac{1}{2}\left(\frac{t^2+1}{t}\right) \end{gathered}$$

Differentiate the Equation $\left(1\right)$

$$\begin{gathered} \Rightarrow secx+\tan x=t \\ \Rightarrow \left(secx\tan x+se{c}^{2}x\right)dx=dt \\ \Rightarrow secx\left(\tan x+secx\right)dx=dt \\ \Rightarrow secxdx=\frac{dt}{\tan x+secx} \\ \Rightarrow secxdx=\frac{dt}{t} \end{gathered}$$

Substitute the above values in the integral.

$$\begin{gathered} I=\int \frac{{\displaystyle \frac{1}{2}}\left(t+{\displaystyle \frac{1}{t}}\right){\displaystyle \frac{dt}{t}}}{t^{{\displaystyle \frac{9}{2}}}} \\ \Rightarrow I=\frac{1}{2}\int \left(t^{-\frac{9}{2}}+t^{\frac{-13}{2}}\right)dt \\ \Rightarrow I=\frac{1}{2}\left(\frac{t^{-\frac{9}{2}+1}}{-{\displaystyle \frac{9}{2}}+1}+\frac{t^{\frac{-13}{2}+1}}{-{\displaystyle \frac{13}{2}}+1}\right)+c \\ \Rightarrow I=\frac{1}{2}\left(\frac{t^{-\frac{7}{2}}}{-{\displaystyle \frac{7}{2}}}+\frac{t^{\frac{-11}{2}}}{-{\displaystyle \frac{11}{2}}}\right)+c \\ \Rightarrow I=\left(-\frac{t^{-\frac{7}{2}}}{7}-\frac{t^{\frac{-11}{2}}}{11}\right)+c \\ \Rightarrow I=\left(-\frac{{\left(secx+\tan x\right)}^{-{\displaystyle \frac{7}{2}}}}{7}-\frac{{\left(secx+\tan x\right)}^{-\frac{11}{2}}}{11}\right)+c \\ \Rightarrow I=\frac{-1}{{\left(secx+\tan x\right)}^{\frac{11}{2}}}\left(\frac{1}{11}+\frac{1}{7}{\left(secx+\tan x\right)}^2\right)+c \end{gathered}$$

Hence, the correct answer is Option C.