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Question Number 173874 by akolade last updated on 20/Jul/22

Answered by a.lgnaoui last updated on 21/Jul/22

$$\mathrm{I}={\int }_{\pi /4}^{15\pi /4}(\sin x\cos {}^{2}x+{3\sin }^{2}x{\cos }^{2}x+{4\sin }^{5}x{\cos }^{2}x+{2\sin }^{3}x)dx=$$$${\cos }^{2}x{\sin }^{2}x(3+{4\sin }^{3}x)+(\sin x{\cos }^{2}x+{2\sin }^{3}x)=$$$$\frac{1}{4}{\left(\sin 2x\right)}^2(3+{4\sin }^{3}x)+\sin x(1+{\sin }^{2}x)=$$$$\frac{3}{4}{\sin }^2\left(2x\right)+{2\sin }^{3}x+\sin x$$$$$$$$\mathrm{I}=\frac{3}{4}\int {\sin }^2\left(2x\right)dx+3\int {\sin }^{3}xdx+\int \sin xdx$$$$\int \sin {}^2\left(2x\right)dx{u}^{\prime }=1\Rightarrow u=xv={\sin }^2\left(2x\right)\Rightarrow v^{\prime }=2\sin \left(2x\right)\cos \left(2x\right)=\sin 4x$$$$\int {\sin }^2\left(2x\right)dx=\left[x{\sin }^2\left(2x\right)\right]-(\int x\sin 4xdx=[-\frac{1}{4}x\cos 4x]+\sin 4x)$$$$\Rightarrow \int {\sin }^{2}2xdx=x{\sin }^{2}2x+\frac{1}{4}\left(x\cos 4x\right)-\sin 4x$$$$\int ({\sin }^{3}x+\sin x)dx=\int \sin x(1+{\sin }^{2}x)=-\cos x(1+{\sin }^{2}x)-\frac{\cos 2x}{2}$$$$\mathrm{I}=\frac{3}{4}{\left[x\sin {}^{2}2x\right]}_{\frac{\pi }{4}}^{\frac{15\pi }{4}}+\frac{3}{16}{\left[x\cos 4x\right]}_{\frac{\pi }{4}}^{\frac{15\pi }{4}}-\frac{3}{4}{\left[\sin 4x\right]}_{\frac{\pi }{4}}^{\frac{15\pi }{4}}$$$$-{\left[\cos x(1+{\sin }^{2}x)\right]}_{\frac{\pi }{4}}^{\frac{15\pi }{4}}+{\left[\frac{\cos 2x}{2}\right]}_{\frac{\pi }{4}}^{\frac{15\pi }{4}}$$$$=\frac{3}{4}\left[(\frac{15\pi }{4}{\sin }^2(-\frac{\pi }{4})-\frac{\pi }{4}{\sin }^2\frac{\pi }{4})\right]+\left[\frac{3}{16}(\frac{15\pi }{4}\cos (-\pi )-\frac{\pi }{4}\cos \pi )\right]-\frac{3}{4}(\sin 15\pi -\sin \pi )$$$$-[\cos \left(\frac{\pi }{4}\right)(1+\sin {}^2\frac{\pi }{4})-\left(\cos \frac{\pi }{4}(1+{\sin }^2\frac{\pi }{4})\right]+(\frac{1}{2}\cos \frac{3\pi }{2}-\frac{1}{2}\cos \frac{\pi }{2})$$$$\mathrm{I}=\frac{-49\pi }{32}$$

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