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Question Number 69603 by Mikael last updated on 25/Sep/19

$$\int x^{3}arcsinxdx$$

Answered by MJS last updated on 25/Sep/19

$$\mathrm{by}\mathrm{parts}$$$$\int u^{\prime }v=uv-\int {uv}^{\prime }$$$$u^{\prime }=x^3\to u=\frac{x^4}{4}$$$$v=\arcsin x\to v^{\prime }=\frac{1}{\sqrt{1-x^2}}$$$$\int x^3\arcsin xdx=\frac{x^4}{4}\arcsin x-\frac{1}{4}\int \frac{x^4}{\sqrt{1-x^2}}dx$$$$\int \frac{x^4}{\sqrt{1-x^2}}dx=$$$$[t=\arcsin x\to dx=\sqrt{1-x^2}dt]$$$$=\int {\sin }^{4}tdt=\frac{1}{8}\int \cos 4tdt-\frac{1}{2}\int \cos 2tdt+\frac{3}{8}\int dt=$$$$=\frac{1}{32}\sin 4t-\frac{1}{4}\sin 2t+\frac{3}{8}t=$$$$=-(\frac{1}{4}{x}^3-\frac{3}{8}x)\sqrt{1-x^2}+\frac{3}{8}\arcsin x$$$$$$$$\int x^3\arcsin xdx=(\frac{1}{4}{x}^4-\frac{3}{32})\arcsin x+\frac{1}{32}x(2x^2+3)\sqrt{1-x^2}+C$$

Commented by Mikaell last updated on 27/Sep/19

$$greatSir.thankyou.$$

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