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Question Number 128664 by bemath last updated on 09/Jan/21

$${\int }_0^{\pi /2}(1-\sin \mathrm{x}+\sin {}^2\mathrm{x}-\sin {}^3\mathrm{x}+\sin {}^4\mathrm{x}-\sin {}^5\mathrm{x}+...)\mathrm{dx}=?$$

Answered by liberty last updated on 09/Jan/21

$$\mathrm{Y}={\int }_0^{\pi /2}\frac{1}{1-(-\sin \mathrm{x})}\mathrm{dx}$$$$\mathrm{Y}={\int }_0^{\pi /2}\frac{1-\sin \mathrm{x}}{1-\sin {}^2\mathrm{x}}\mathrm{dx}={\int }_0^{\pi /2}\left(\frac{1-\sin \mathrm{x}}{\cos {}^2\mathrm{x}}\right)\mathrm{dx}$$$$\mathrm{Y}={\int }_0^{\pi /2}(\sec {}^2\mathrm{x}-\frac{\sin \mathrm{x}}{\cos {}^2\mathrm{x}})\mathrm{dx}$$$$\mathrm{Y}=\underset{\mathrm{b}\to {\left(\frac{\pi }{2}\right)}^{-}}{\lim }(\tan \mathrm{x}-\frac{1}{\cos \mathrm{x}}){\mid }_0^{\mathrm{b}}$$$$\mathrm{Y}=\underset{\mathrm{b}\to {\left(\frac{\pi }{2}\right)}^{-}}{\lim }\left(\frac{\sin \mathrm{b}-1}{\cos \mathrm{b}}\right)-(0-1)$$$$\mathrm{Y}=1+\underset{\mathrm{b}\to {\left(\frac{\pi }{2}\right)}^{-}}{\lim }\left(\frac{\cos \mathrm{b}}{-\sin \mathrm{b}}\right)=1+0=1.$$

Answered by Dwaipayan Shikari last updated on 09/Jan/21

$${\int }_0^{\frac{\pi }{2}}\frac{1}{1+sinx}dx$$$$=2{\int }_0^1\frac{1}{1+\frac{2t}{1+t^2}}.\frac{1}{1+t^2}dtt=tan\frac{x}{2}$$$$=-{\left[\frac{2}{(1+t)}\right]}_0^1=1$$

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