# Chapter 5 - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises - Page 338: 45

$$\int^{-\sqrt2/2}_{-1}\frac{dy}{y\sqrt{4y^2-1}}=-\frac{\pi}{12}$$

#### Work Step by Step

$$A=\int^{-\sqrt2/2}_{-1}\frac{dy}{y\sqrt{4y^2-1}}=\int^{-\sqrt2/2}_{-1}\frac{ydy}{y^2\sqrt{4y^2-1}}$$ We set $u=\sqrt{4y^2-1}$, which means $4y^2-1=u^2$ and $y^2=\frac{u^2+1}{4}$ $$du=\frac{(4y^2-1)'}{2\sqrt{4y^2-1}}dy=\frac{8y}{2\sqrt{4y^2-1}}dy$$ $$du=\frac{4y}{\sqrt{4y^2-1}}dy$$ $$\frac{ydy}{\sqrt{4y^2-1}}=\frac{1}{4}du$$ - For $y=-\sqrt2/2$, we have $u=\sqrt{4(-\sqrt2/2)^2-1}=\sqrt{4\times1/2-1}=1$ - For $y=-1$, we have $u=\sqrt{4(-1)^2-1}=\sqrt3$ Therefore, $$A=\frac{1}{4}\int^{1}_{\sqrt3}\frac{du}{\frac{u^2+1}{4}}=\frac{1}{4}\int^{1}_{\sqrt3}\frac{4du}{u^2+1}=\int^{1}_{\sqrt3}\frac{du}{u^2+1}$$ $$A=\tan^{-1} u\Big]^{1}_{\sqrt3}$$ $$A=\tan^{-1}1-\tan^{-1}\sqrt3$$ $$A=\frac{\pi}{4}-\frac{\pi}{3}=-\frac{\pi}{12}$$

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