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Question Number 57750 by maxmathsup by imad last updated on 11/Apr/19

$$find\int x^2\sqrt{25-x^2}dx$$

Answered by MJS last updated on 11/Apr/19

$$\int x^2\sqrt{25-x^2}dx=$$$$[t={\sin }^{-1}\frac{x}{5}\to dx=5\cos tdt]$$$$=625\int {\sin }^{2}t{\cos }^{2}tdt=\frac{625}{8}\int (1-\cos 4t)dt=$$$$=\frac{625}{8}t-\frac{625}{32}\sin 4t=\frac{625}{8}{\sin }^{-1}\frac{x}{5}-\frac{625}{32}\sin \left({4\sin }^{-1}\frac{x}{5}\right)=$$$$=\frac{625}{8}{\sin }^{-1}\frac{x}{5}+\frac{1}{8}x(2x^2-25)\sqrt{25-x^2}+C$$

Commented by maxmathsup by imad last updated on 11/Apr/19

$$thankssirmjs$$

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