Question

$\int \frac{1-\sin x}{1+\sin x}dx=$

• A. $2\tan x-2\sec x-x+c$
• B. $2\tan x-\sec x-x+c$
• C. $\tan x+2\sec x+x+c$
• D. $\tan x-2\sec x+x+c$

Answer 1

Answer: (A)

$2\tan x-2\sec x-x+c$

$\int \frac{1-\sin x}{1+\sin x}dx=\int \frac{(1-\sin x)(1-\sin x)}{(1+\sin x)(1-\sin x)}\cdot dx$

$=\int \frac{(1-\sin x{)}^2}{{\cos }^{2}x}\cdot dx$

$=\int \frac{1+{\sin }^{2}x-2\sin x}{{\cos }^{2}x}\cdot dx$

$\int {\sec }^{2}x\cdot dx+\int {\tan }^{2}x\cdot dx-2\int \tan x\cdot \sec xdx$

$\int {\sec }^{2}x\cdot dx+\int ({\sec }^{2}x-1)\cdot dx-2\int \tan x\cdot \sec xdx$

$=2\int {\sec }^{2}x\cdot dx-\int dx-2\int \tan x\sec xdx$

$=2\tan x-x-2\sec x+c$

$\int \frac{1-\sin x}{1+\sin x}\cdot dx=2\tan x-x-2\sec x+c$