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Question Number 201289 by sonukgindia last updated on 03/Dec/23

Commented by mr W last updated on 03/Dec/23

$$sir:$$$$whydoyouignoreallquestions$$$$otherpeopleareaskingyou?$$

Answered by BaliramKumar last updated on 04/Dec/23

$${\int }_0^{\frac{\pi }{2}}{\sin }^2\mathrm{xdx}={\int }_0^{\frac{\pi }{2}}\sqrt{1-{\cos }^2\mathrm{x}}\mathrm{sinxdx}=$$$$$$$$\mathrm{put}\mathrm{cosx}=\mathrm{t}-\mathrm{sinxdx}=\mathrm{dt}$$$$\mathrm{x}=\frac{\pi }{2}\to \mathrm{t}=0$$$$\mathrm{x}=0\to \mathrm{t}=1$$$$-{\int }_1^0\sqrt{1-{\mathrm{t}}^2}\mathrm{dt}={\int }_0^1\sqrt{1-{\mathrm{t}}^2}\mathrm{dt}$$$${[\frac{\mathrm{t}}{2}\sqrt{1-{\mathrm{t}}^2}+\frac{1}{2}{\sin }^{-1}\mathrm{t}]}_0^1=\frac{\pi }{4}$$$$$$

Answered by Calculusboy last updated on 03/Dec/23

$$\mathbf{S}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}:\mathit{b}\mathit{y}\mathit{u}\mathit{s}\mathit{i}\mathit{n}\mathit{g}\mathit{w}\mathit{a}\mathit{l}\mathit{l}\mathit{i}\mathit{s}\mathit{f}\mathit{o}\mathit{r}\mathit{m}\mathit{u}\mathit{l}\mathit{a}$$$${\mathit{I}}_{\mathit{n}}=\frac{\mathit{n}-1}{\mathit{n}}\cdot \frac{\mathit{n}-3}{\mathit{n}-2}\cdot \cdot$$$$\mathit{s}\mathit{i}\mathit{n}\mathit{c}\mathit{e}\mathit{t}\mathit{h}\mathit{e}\mathit{p}\mathit{o}\mathit{w}\mathit{e}\mathit{r}\mathit{i}\mathit{s}\mathit{e}\mathit{v}\mathit{e}\mathit{n}$$$${\mathit{I}}_{\mathit{n}}=\frac{2-1}{2}=\frac{1}{2}\times \frac{\mathit{\pi }}{2}=\frac{\mathit{\pi }}{4}$$$$\therefore {\int }_0^{\frac{\mathit{\pi }}{2}}{\mathit{s}\mathit{i}\mathit{n}}^2\mathit{x}\mathit{d}\mathit{x}=\frac{\mathit{\pi }}{4}$$

Answered by mr W last updated on 04/Dec/23

$$={\int }_0^{\frac{\pi }{2}}\frac{1-\cos 2x}{2}dx$$$$=\frac{1}{2}{[x-\frac{\sin 2x}{2}]}_0^{\frac{\pi }{2}}$$$$=\frac{\pi }{4}$$

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