Question

How to solve this integration problem? the answer should be -π/12

Answer 1

$\displaystyle\int_{-1}^{-\sqrt2/2}\frac{\mathrm dy}{y\sqrt{4y^2-1}}$

Try setting $y=\dfrac12\sec t$, so that $\mathrm dy=\dfrac12\sec t\tan t\,\mathrm dt$. The interval then changes from $-1\le y\le-\dfrac{\sqrt2}2$ to $\dfrac{2\pi}3\le t\le \dfrac{3\pi}4$.

Now, the integral is

$\displaystyle \int_{2\pi/3}^{3\pi/4} \frac{\dfrac12\sec t\tan t}{\dfrac12\sec t\sqrt{4\left(\dfrac12\sec t\right)^2-1}}\,\mathrm dt=\int_{2\pi/3}^{3\pi/4}\frac{\tan t}{\sqrt{\sec^2t-1}}\,\mathrm dt$

The Pythagorean identity lets you reduce the denominator:

$\sqrt{\sec^2t-1}=\sqrt{\tan^2t}=|\tan t|$

Since $\tan t$ for the given interval, you have $|\tan t|=-\tan t$, which means the integral is equal to

$\displaystyle\int_{2\pi/3}^{3\pi/4}\frac{\tan t}{-\tan t}\,\mathrm dt=-\int_{2\pi/3}^{3\pi/4}\mathrm dt=-\dfrac\pi{12}$