Question

$\int (\sqrt{\sin x}+\sqrt{\cos x}{)}^{-4}dx=$

• A. $\frac{2}{(\sqrt{\tan x}+1{)}^2}[\frac{1}{3(\sqrt{\tan x}+1)}-\frac{1}{2}]+c$
• B. $\frac{2}{(\sqrt{\tan x}+1{)}^2}[\frac{1}{3(\sqrt{\tan x}+1)}+\frac{1}{2}]+c$
• C. $\frac{2}{(\sqrt{\tan x}+1{)}^2}[\frac{1}{\sqrt{3}(\sqrt{\tan x}+1)}+\frac{1}{2}]+c$
• D. $\frac{2}{(\sqrt{\tan x}+1{)}^2}[\frac{1}{\sqrt{3}(\sqrt{\tan x}+1)}-\frac{1}{2}]+c$

Answer 1

The correct option is A $\frac{2}{(\sqrt{\tan x}+1{)}^2}[\frac{1}{3(\sqrt{\tan x}+1)}-\frac{1}{2}]+c$

$I=\int (\sqrt{\sin x}+\sqrt{\cos x}{)}^{-4}dx=\int \frac{dx}{\left[\sqrt{\cos x}\right(\sqrt{\tan x}+1){]}^4}=\int \frac{dx}{{\cos }^{2}x(\sqrt{\tan x}+1{)}^4}=\int \frac{{\sec }^{2}xdx}{(\sqrt{\tan x}+1{)}^4}$
Put $\sqrt{\tan x}+1=y\Rightarrow \frac{1}{2\sqrt{\tan x}}{\sec }^{2}xdx=dy\Rightarrow {\sec }^{2}xdx=2\sqrt{\tan x}dy=2(y-1)dyI=\int \frac{1}{y^4}.2(y-1)dy=2\int (\frac{1}{y^3}-\frac{1}{y^4})dy=2[\frac{-1}{2y^2}+\frac{1}{3y^3}]+c=\frac{2}{y^2}[\frac{1}{3y}-\frac{1}{2}]+c=\frac{2}{(\sqrt{\tan x}+1{)}^2}[\frac{1}{3(\sqrt{\tan x}+1)}-\frac{1}{2}]+c$