Question

If $I={\int }_{\pi /3}^{2\pi /3}\frac{x}{1+\sin x}dx$ then $I$ equals

• A. $2\pi$
• B. $(2+\sqrt{3})\pi$
• C. $(2-\sqrt{3})\pi$
• D. $0$

Answer 1

The correct option is C $(2-\sqrt{3})\pi$

Using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx,$
$I={\int }_{\pi /3}^{2\pi /3}\frac{x}{1+\sin x}dx$
We can write
$I={\int }_{\pi /3}^{2\pi /3}\frac{\pi -x}{1+\sin x}dx$
Adding we get
$I={\int }_{\pi /3}^{2\pi /3}\frac{\pi .dx}{1+\sin x}$
Substitute $\tan \left(x/2\right)=t$ so that,
$2I={\int }_{1/\sqrt{3}}^{\sqrt{3}}\frac{2\pi \cdot dt(1+t^2)}{1+2t(1+t^2)}$
$\Rightarrow I={\int }_{1/\sqrt{3}}^{\sqrt{3}}\frac{\pi .dt}{(1+t{)}^2}={\frac{-\pi }{1+t}]}_{1/\sqrt{3}}^{\sqrt{3}}$
$=\pi [-\frac{1}{1+\sqrt{3}}+\frac{\sqrt{3}}{\sqrt{3}+1}]=\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)\left(\frac{\sqrt{3}-1}{\sqrt{3}-1}\right)\pi$
$=(2-\sqrt{3})\pi$