Question

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

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

Answer 1

Answer: (C)

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

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

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

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

Now $(\sec x\tan x+{\sec }^{2}x)dx=dt$

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

$\sec xdx=\frac{dt}{t},\frac{1}{2}(t+\frac{1}{t})=\sec x$

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

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

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

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

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

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