Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 73429 by Henri Boucatchou last updated on 12/Nov/19

$$Solve:\int \frac{{\left[{cos}^{-1}x\sqrt{1-x^2}\right]}^{-1}}{{log}_e[2+\frac{sin\left(2x\sqrt{1-x^2}\right)}{\pi }]}dx$$$$Evaluate{\int }_{-\pi /2}^{\pi /2}{sin}^2{xcos}^{2}x(cosx+sinx)dx$$

Commented by MJS last updated on 12/Nov/19

$$\int {\sin }^{2}x{\cos }^{2}x(\cos x+\sin x)dx=$$$$=\int {\sin }^{3}x{\cos }^{2}xdx+\int {\sin }^{2}x{\cos }^{3}xdx=$$$$=-\frac{\int \sin 5xdx}{16}+\frac{\int \sin 3xdx}{16}+\frac{\int \sin xdx}{8}-$$$$-\frac{\int \cos 5xdx}{16}-\frac{\int \cos 3xdx}{16}+\frac{\int \cos xdx}{8}=$$$$=\frac{\cos 5x}{80}-\frac{\cos 3x}{48}-\frac{\cos x}{8}-\frac{\sin 5x}{80}-\frac{\sin 3x}{48}+\frac{\sin x}{8}+C$$$$\underset{-\frac{\pi }{2}}{\stackrel{\frac{\pi }{2}}{\int }}{\sin }^{2}x{\cos }^{2}x(\cos x+\sin x)dx=\frac{4}{15}$$

Commented by MJS last updated on 12/Nov/19

$$\mathrm{please}\mathrm{check}\mathrm{the}\mathrm{first}\mathrm{one},\mathrm{is}\mathrm{it}\sin \mathrm{or}{\sin }^{-1}?$$

Commented by MJS last updated on 12/Nov/19

$$...\mathrm{anyway}\mathrm{I}\mathrm{cannot}\mathrm{solve}\mathrm{it}$$

Answered by MJS last updated on 12/Nov/19

$$\int {\sin }^{2}x{\cos }^{2}x(\cos x+\sin x)dx=$$$$=\int {\sin }^{3}x{\cos }^{2}xdx+\int {\sin }^{2}x{\cos }^{3}xdx=$$$$[u=\cos x\to dx=-\frac{du}{\sin x};v=\sin x\to dx=\frac{dv}{\cos x}]$$$$=\int u^4-u^{2}du+\int v^2-v^{4}dv=$$$$=\frac{u^5}{5}-\frac{u^3}{3}+\frac{v^3}{3}-\frac{v^5}{5}=$$$$=\frac{{\cos }^{5}x}{5}-\frac{{\cos }^{3}x}{3}+\frac{{\sin }^{3}x}{3}-\frac{{\sin }^{5}x}{5}+C$$$$\underset{-\frac{\pi }{2}}{\stackrel{\frac{\pi }{2}}{\int }}{\sin }^{2}x{\cos }^{2}x(\cos x+\sin x)dx=\frac{4}{15}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com