Question

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

• A. $(\sec x+\tan x{)}^{-\frac{13}{2}}(\frac{1}{7}+\frac{1}{11}(\sec x+\tan x{)}^{\frac{13}{7}})+k$
• B. $(\sec x-\tan x{)}^{-\frac{7}{2}}(\frac{1}{7}-\frac{1}{11}(\sec x+\tan x{)}^{\frac{11}{7}})+k$
• C. $-(\sec x+\tan x{)}^{-\frac{7}{2}}(\frac{1}{7}+\frac{1}{11}(\sec x+\tan x{)}^{\frac{11}{7}})+k$
• D. $-(\sec x-\tan x{)}^{-\frac{7}{2}}(\frac{1}{7}-\frac{1}{11}(\sec x+\tan x{)}^{\frac{11}{7}})+k$

Answer 1

The correct option is C

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

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

Let $\sec x+\tan x=t$

$\Rightarrow \sec x(\sec x+\tan x)dx=dt$

$\Rightarrow \sec xdx=\frac{dt}{t}$

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

$\Rightarrow \sec x=\frac{1}{2}(t+\frac{1}{t})$

Integration becomes $I=\frac{1}{2}\int \frac{(t+\frac{1}{t})}{t^{\frac{9}{2}}}\frac{dt}{t}=\frac{1}{2}\int (t^{-\frac{9}{2}}+t^{\frac{-13}{2}})dt$

$=\frac{1}{2}[\frac{t^{-\frac{9}{2}+1}}{-\frac{9}{2}+1}+\frac{t^{-\frac{13}{2}+1}}{-\frac{13}{2}+1}]+k$

$=\frac{1}{2}[\frac{t^{-\frac{7}{2}}}{-\frac{7}{2}}+\frac{t^{-\frac{11}{2}}}{-\frac{11}{2}}]+k$

$=-\frac{1}{7}{t}^{-\frac{7}{2}}-\frac{1}{11}{t}^{-\frac{11}{2}}+k$

$=-t^{-\frac{7}{2}}(\frac{1}{7}+\frac{1}{11}{t}^{\frac{11}{7}})+k$

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