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

The correct option is C $-\frac{1}{(\sec x+\tan x{)}^{11/2}}\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x{)}^2\}+K$

Let $\sec x+\tan x=t$
$\Rightarrow \sec x(\sec x+\tan x)dx=dt$
$\Rightarrow dx=\frac{2}{1+t^2}dt$
Now, $t=\int \frac{{\left(\frac{t+\frac{1}{t}}{2}\right)}^2\frac{2}{1+t^2}dt}{t^{9/2}}$
$=\frac{1}{2}\int (t^{-9/2}+t^{-13/2})dt$
$=\frac{1}{2}\{\frac{-2}{7}{t}^{\frac{-7}{2}}-\frac{2}{11}{t}^{-\frac{11}{2}}\}+K$
$=-t^{-\frac{11}{2}}\{\frac{1}{7}{t}^2+\frac{1}{11}\}+K$
$=-\frac{1}{(\sec x+\tan x{)}^{11/2}}\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x{)}^2\}+K$