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Question Number 155421 by SANOGO last updated on 30/Sep/21

Answered by puissant last updated on 30/Sep/21

$$Q={\int }_0^{\frac{\pi }{2}}\frac{sint}{1+{cos}^{2}t}dt=-{\int }_0^{\frac{\pi }{2}}\frac{-sint}{1+{cos}^{2}t}dt$$$$=-{\left[arctan\left(cost\right)\right]}_0^{\frac{\pi }{2}}=-(0-arctan\left(1\right))$$$$$$$$\therefore \because Q=\frac{\pi }{4}..$$

Commented by SANOGO last updated on 30/Sep/21

$$mercibien$$

Answered by ArielVyny last updated on 30/Sep/21

$$u=tan\left(\frac{t}{2}\right)\to du=\frac{1}{2}(1+u^2)dt\to dt=\frac{2}{1+u^2}du$$$${\int }_0^1\frac{\frac{2u}{1+u^2}}{1+{\left(\frac{1-u^2}{1+u^2}\right)}^2}.\frac{2}{1+u^2}du={\int }_0^1\frac{2u}{1+u^2}.\frac{2}{(1+u^2)+\frac{{(1-u^2)}^2}{1+u^2}}du$$$${\int }_0^1\frac{4u}{{(1+u^2)}^2+{(1-u^2)}^2}du=4{\int }_0^1\frac{u}{u^4+2u^2+1+u^4-2u^2+1}$$$$4{\int }_0^1\frac{u}{2u^4+2}du=2{\int }_0^1\frac{u}{u^4+1}du$$$$u^2=x\to 2udu=dx$$$${\int }_0^1\frac{dx}{x^2+1}={\int }_0^{\frac{\pi }{2}}\frac{sint}{1+{cos}^{2}t}dt=\frac{\pi }{4}$$

Commented by SANOGO last updated on 30/Sep/21

$$mercibien$$

Answered by physicstutes last updated on 30/Sep/21

$$\mathrm{let}x=\cos t\Rightarrow dx=-\sin tdt$$$$\overline{)\begin{array}{ccc}t& 0& \frac{\pi }{2}\\ x& 1& 0\end{array}}$$$$\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{\sin t}{1+{\cos }^{2}t}dt=\underset{1}{\stackrel{0}{\int }}\frac{\sin t}{1+x^2}(-\frac{dx}{\sin t})=\underset{0}{\stackrel{1}{\int }}\frac{dx}{1+x^2}$$$${={\tan }^{-1}\left(x\right)]}_0^1=\frac{\pi }{4}$$

Commented by SANOGO last updated on 30/Sep/21

$$mercibien$$

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