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Question Number 11436 by @ANTARES_VY last updated on 26/Mar/17

$$$$$$\underset{0}{\stackrel{\frac{\mathit{\pi }}{4}}{\int }}\mathbf{sinx}\times {\mathbf{cos}}^7\mathbf{x}\mathbf{dx}.$$$$\mathbf{solves}...$$

Commented by FilupS last updated on 26/Mar/17

$$\mathrm{let}u=\cos \left(x\right)$$$$\therefore du=-\sin \left(x\right)dx$$$$$$$$\therefore {\int }_0^{\pi /4}\sin \left(x\right){\cos }^7\left(x\right)dx=-{\int }_{x=0}^{x=\pi /4}{u}^{7}du$$$$-{\int }_{x=0}^{x=\pi /4}{u}^{7}du=-{\left[\frac{1}{8}{u}^8\right]}_{x=0}^{x=\pi /4}$$$$=-\frac{1}{8}{\left[{\cos }^8\left(x\right)\right]}_{x=0}^{x=\pi /4}$$$$=-\frac{1}{8}({\cos }^8\left(\frac{\pi }{4}\right)-{\cos }^8\left(0\right))$$$$=-\frac{1}{8}({\left(\frac{1}{\sqrt{2}}\right)}^8-1)$$$$=-\frac{1}{8}(2^{-\frac{1}{2}\times 8}-1)$$$$=-\frac{1}{8}(2^{-4}-1)$$$$=-\frac{1}{2^3}(\frac{1}{2^4}-1)$$$$=-(\frac{1}{2^7}-\frac{1}{2^3})$$$$=-(\frac{1}{2^7}-\frac{2^4}{2^7})$$$$=-\left(\frac{1-2^4}{2^7}\right)$$$$=\frac{2^4-1}{2^7}$$$$=\frac{15}{128}$$

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