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 119784 by bemath last updated on 27/Oct/20

$$\int \frac{dx}{\sqrt{{(4x-x^2)}^3}}$$$$$$

Answered by bobhans last updated on 27/Oct/20

$$\int \frac{dx}{{(4-{(2-x)}^2)}^{\frac{3}{2}}};[let2-x=2\sin t]$$$$dx=-2\cos tdt]$$$$\int \frac{-2\cos tdt}{{(4-4\sin {}^{2}t)}^{\frac{3}{2}}}=\int \frac{-\cos tdt}{4\cos {}^{3}t}$$$$=-\frac{1}{4}\int \frac{1}{\cos {}^{2}t}dt=-\frac{1}{4}\tan t+c$$$$=-\frac{2-x}{4\sqrt{4x-x^2}}+c=\frac{x-2}{4\sqrt{4x-x^2}}+c$$

Answered by MJS_new last updated on 28/Oct/20

$$\int \frac{dx}{x^{3/2}{(4-x)}^{3/2}}=$$$$[t=\frac{\sqrt{x}}{\sqrt{4-x}}\to dx=\frac{\sqrt{x{(4-x)}^3}}{2}dt]$$$$=\frac{1}{8}\int 1+\frac{1}{t^2}dt=\frac{t}{8}-\frac{1}{8t}=\frac{x-2}{4\sqrt{x(4-x)}}+C$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com