Question

If $F\left(x\right)=\int \frac{x+\sin x}{1+\cos x}dx$ and $F\left(0\right)=0,$ then the value of $F\left(\pi /2\right)$ is

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{4}$
• D. $\pi$

Answer 1

The correct option is A $\frac{\pi }{2}$

$F\left(x\right)=\int \frac{x+sinx}{1+cosx}dx$ and F(0)= 0, then the value

$F\left(\frac{\pi }{2}\right)$ is

$F\left(x\right)=\int \frac{x+2sin\frac{x}{2}cos\frac{x}{2}}{2co{s}^2\frac{x}{2}}dx$

$=\int \frac{1}{2}xse{c}^2\frac{x}{2}+tan\frac{x}{2}dx$

$=\frac{1}{2}[x\int se{c}^2\frac{x}{2}dx-\int (\frac{d}{dx}x\int se{c}^2\frac{x}{2}dx\left)dx\right]+\int tan\frac{x}{2}dx$

$=\frac{x}{2}\frac{tan\frac{x}{2}}{\frac{\pi }{2}}-\frac{1}{2}\int \frac{tan\frac{x}{2}}{\frac{1}{2}}dx+\int tan\frac{x}{2}dx$

$\therefore F\left(x\right)=xtan\frac{x}{2}+c$

$\therefore F\left(0\right)=\left(0\right)tan\frac{0}{2}+c$

$\therefore c=0$

$\therefore F\left(x\right)=xtan\frac{x}{2}$

$\therefore F\left(\frac{\pi }{2}\right)=\frac{\pi }{2}tan\frac{\pi }{4}=\frac{\pi }{2}$