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Question Number 193411 by cortano12 last updated on 13/Jun/23

Commented by BaliramKumar last updated on 13/Jun/23

$$\mathrm{put}x=cos2\theta$$

Answered by Frix last updated on 13/Jun/23

$$\underset{0}{\stackrel{1}{\int }}\sqrt{\frac{1-x}{1+x}}dx\stackrel{t=\sqrt{\frac{1+x}{1-x}}}{=}$$$$=4\underset{1}{\stackrel{\mathrm{\infty }}{\int }}\frac{dt}{{(t^2+1)}^2}={[\frac{2t}{t^2+1}+{2\tan }^{-1}t]}_1^{\mathrm{\infty }}=$$$$=\frac{\pi }{2}-1$$

Answered by Subhi last updated on 13/Jun/23

$$putx=cos\left(\theta \right)\Rrightarrow dx=-sin\left(\theta \right).d\theta$$$${\int }_0^1\sqrt{\frac{1-cos\left(\theta \right)}{1+cos\left(\theta \right)}}.-sin\left(\theta \right)d\theta$$$$-{\int }_0^{1}tan\left(\frac{\theta }{2}\right).sin\left(\theta \right)d\theta \Rrightarrow -2\int \frac{sin\left(\frac{\theta }{2}\right)}{cos\left(\frac{\theta }{2}\right)}.sin\left(\frac{\theta }{2}\right).cos\left(\frac{\theta }{2}\right)d\theta$$$$-2\int {sin}^2\left(\frac{\theta }{2}\right)d\theta \Rrightarrow -\int 1-cos\left(\theta \right)d\theta \Rrightarrow sin\left(\theta \right)-\theta$$$$x=\sqrt{1-{sin}^2\left(\theta \right)}\Rrightarrow sin\left(\theta \right)=\sqrt{1-x^2}$$$${[\sqrt{1-x^2}-{cos}^{-1}\left(x\right)]}_0^1=0-(1-\frac{\pi }{2})=\frac{\pi }{2}-1$$$$$$

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