$\displaystyle\int_{\pi/6}^{\pi/3} \dfrac{dx}{1+\sqrt{ an x}}$ equals

Question

$\displaystyle\int_{\pi/6}^{\pi/3} \dfrac{dx}{1+\sqrt{	an x}}$ equals

Accepted Answer

C. $\dfrac{\pi}{12}$

1. Write $I=\int_{\pi/6}^{\pi/3}\dfrac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx$.
2. By $\mathbf{P_3}$ with $a+b=	frac{\pi}{2}$, replace $x$: numerator becomes $\sqrt{\sin x}$.
3. Adding: $2I=\int_{\pi/6}^{\pi/3} dx=\dfrac{\pi}{3}-\dfrac{\pi}{6}=\dfrac{\pi}{6}$.
4. Hence $I=\dfrac{\pi}{12}$.
_Source: syllabus Mathematics Ch 7 "Integrals", p.345_

Answer

$\dfrac{\pi}{6}$

Answer

$\dfrac{\pi}{3}$

Answer

$\dfrac{\pi}{4}$

$\displaystyle\int_{\pi/6}^{\pi/3} \dfrac{dx}{1+\sqrt{\tan x}}$ equals

Keep practising Straight Lines and Conic Sections