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Question Number 48474 by mondodotto@gmail.com last updated on 24/Nov/18

$$\mathbf{evaluate}{\int }_0^{\pi }{\mathbf{sin}}^2\mathit{x}\mathbf{d}\mathit{x}$$

Answered by hassentimol last updated on 24/Nov/18

Commented by hassentimol last updated on 24/Nov/18

$$\mathrm{I}\mathrm{was}\mathrm{in}\mathrm{degrees}...$$

Answered by tanmay.chaudhury50@gmail.com last updated on 24/Nov/18

$${\int }_0^{\pi }\frac{1-cos2x}{2}dx$$$$=\frac{1}{2}\mid x-\frac{sin2x}{2}{\mid }_0^{\pi }$$$$=\frac{1}{2}\times \pi$$$$or{\int }_0^{\pi }{sin}^{2}xdx{\int }_0^{2a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dxwhenf\left(x\right)=f(2a-x)$$$$=2{\int }_0^{\frac{\pi }{2}}{sin}^{2}xdx$$$$=2\times \frac{1}{2}\mid x-\frac{sin2x}{2}{\mid }_0^{\frac{\pi }{2}}$$$$=2\times \frac{1}{2}\times \frac{\pi }{2}=\frac{\pi }{2}$$

Commented by mondodotto@gmail.com last updated on 26/Nov/18

$$\mathrm{sir}\mathrm{it}\mathrm{is}\sin {\mathrm{x}}^2\mathrm{not}{\sin }^2\mathrm{x}$$

Commented by tanmay.chaudhury50@gmail.com last updated on 26/Nov/18

$$ok...butinquestionitis{sin}^{2}x$$

Commented by hassentimol last updated on 26/Nov/18

$$\mathrm{Well}...$$$$\sin {\left(x\right)}^2={\left(\sin \left(x\right)\right)}^2={\sin }^2\left(x\right)\mathrm{by}\mathrm{definition}$$

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