Question

$\int \frac{1}{1+\sin x}dx=\tan (\frac{x}{2}+a)+C,$ then

• A. $a=-\frac{-\pi }{4},CϵR$
• B. $a=-\frac{\pi }{4},CϵR$
• C. $a=-\frac{5\pi }{4},CϵR$
• D. $a=-\frac{\pi }{3},CϵR$

Answer 1

The correct option is A $a=-\frac{-\pi }{4},CϵR$

$\int \frac{1}{1+sinx}dx=tan(\frac{x}{2}+a)+C,$ then

By differentiating,

$\frac{1}{1+sinxdx}={sec}^2(\frac{x}{2}+a)\left(\frac{1}{2}\right)+0$

$\frac{1}{1+sinx}=\frac{1}{2{cos}^2(\frac{x}{2}+a)}$

$2{cos}^2\theta =1+cos2\theta$

$\frac{1}{1+sinx}=\frac{1}{1+cos2(\frac{x}{2}+a)}$

$\frac{1}{1+sinx}=\frac{1}{1+cos(x+2a)}$

$1+sinx=1+cos(x+2a)$

$sinx=cos(\frac{\pi }{2}+x)$

$x+\frac{\pi }{2}=(x+2a)$

$\frac{\pi }{2}=2a$

$a=\frac{\pi }{4}$