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Question Number 83266 by 09658867628 last updated on 29/Feb/20

$$\int {\sin }^{10}\mathrm{\Theta }\cos \mathrm{\Theta }d\mathrm{\Theta }$$

Commented by Tony Lin last updated on 29/Feb/20

$$letsin\theta =t,dt=cos\theta d\theta$$$$\int t^{10}dt$$$$=\frac{t^{11}}{11}+c$$$$=\frac{{\left(sin\theta \right)}^{11}}{11}+c$$

Answered by Rio Michael last updated on 29/Feb/20

$$----------$$$$\mathbf{solution}$$$$\int \sin {}^{10}\theta \cos \theta d\theta$$$$\mathrm{let}\sin \theta =u\Rightarrow \frac{du}{d\theta }=\cos \theta$$$$\therefore d\theta =\frac{du}{\cos \theta }$$$$\Rightarrow \int {\sin }^{10}\theta \cos \theta d\theta =\int u^{10}\cos \theta \frac{du}{\cos \theta }$$$$=\int u^{10}du=\frac{u^{11}}{11}+k$$$$\Rightarrow \int {\sin }^{10}\theta \cos \theta d\theta =\frac{{\sin }^{11}\theta }{11}+k$$

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