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Question Number 183073 by Mastermind last updated on 19/Dec/22

$$\int {\sin }^2\mathrm{x}\mathrm{dx}$$$$$$$$\mathrm{Integrate}\mathrm{this}\mathrm{question}$$

Answered by Ml last updated on 20/Dec/22

$$\int {\sin }^2\mathrm{xdx}=?$$$$\mathrm{note}\therefore {\sin }^2\mathrm{x}=\frac{1-\cos 2\mathrm{x}}{2}$$$$\int \frac{1-\cos 2\mathrm{x}}{2}\mathrm{dx}=\frac{1}{2}\int (1-\cos 2\mathrm{x})\mathrm{dx}$$$$\Rightarrow \frac{1}{2}\mathrm{x}-\frac{1}{4}\sin 2\mathrm{x}+\mathrm{C},\mathrm{C}\in \mathbb{R}$$$$\mathrm{Eng}-★$$

Commented by Mastermind last updated on 19/Dec/22

$${\mathrm{That}}^{\prime }\mathrm{s}\mathrm{good},\mathrm{you}\mathrm{got}\mathrm{it}\mathrm{correct}!$$

Answered by BaliramKumar last updated on 20/Dec/22

$$\int {sin}^{2}xdx$$$$\int \sqrt{{sin}^{2}x}\cdot sinxdx$$$$\int \sqrt{1-{cos}^{2}x}\cdot sinxdx$$$$letcosx=t-sinxdx=dt$$$$-\int \sqrt{1-t^2}dt$$$$..........$$

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