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 57748 by maxmathsup by imad last updated on 11/Apr/19

$$find\int x^2\sqrt{4+x^2}dx$$

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

$$letI=\int x^2\sqrt{4+x^2}dxchangementx=2sh\left(t\right)give$$$$I=\int 4{sh}^2\left(t\right)2ch\left(t\right)2ch\left(t\right)dt=16\int {sh}^2\left(t\right){ch}^2\left(t\right)dt$$$$=16\int {\left(\frac{1}{2}sh\left(2t\right)\right)}^{2}dt=4\int {sh}^2\left(2t\right)dt=4\int \frac{ch\left(4t\right)-1}{2}dt$$$$=2\int ch\left(4t\right)dt-2t+c=\frac{1}{2}sh\left(4t\right)-2t+c=sh\left(2t\right)ch\left(2t\right)-2t+c$$$$=2sh\left(t\right)ch\left(t\right)(2{ch}^{2}t-1)-2t+c(lookthat{ch}^{2}t+{sh}^{2}t=ch\left(2t\right))$$$$=2sh\left(t\right)ch\left(t\right)(2(1+{sh}^{2}t)-1)-2t+c$$$$=2\frac{x}{2}\sqrt{1+{\left(\frac{x}{2}\right)}^2}(1+2{\left(\frac{x}{2}\right)}^2)-2argsh\left(\frac{x}{2}\right)+c$$$$=x\sqrt{1+\frac{x^2}{4}}(1+\frac{x^2}{2})-2ln(\frac{x}{2}+\sqrt{1+{\left(\frac{x}{2}\right)}^2})+c$$$$$$

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

$$\int x^2\sqrt{4+x^2}dx=\frac{x}{4}(x^2+2)\sqrt{x^2+4}-2ln\left(\frac{x+\sqrt{x^2+4}}{2}\right)+c$$$$=\frac{x}{4}(x^2+2)\sqrt{x^2+4}-2ln(x+\sqrt{x^2+4})+C.$$

Answered by MJS last updated on 11/Apr/19

$$\int x^2\sqrt{x^2+4}dx=$$$$[t={\sinh }^{-1}\frac{x}{2}\to dx=2\cosh tdt]$$$$=16\int {\sinh }^{2}t{\cosh }^{2}tdt=-2\int (1-\cosh 4t)dt=$$$$=-2\int dt+2\int \cosh 4tdt=-2t+\frac{1}{2}\sinh 4t=$$$$=-{2\sinh }^{-1}\frac{x}{2}+\frac{1}{2}\sinh \left({4\sinh }^{-1}\frac{x}{2}\right)+C$$

Commented by MJS last updated on 11/Apr/19

$$=-{2\sinh }^{-1}\frac{x}{2}+\frac{1}{4}x(x^2+2)\sqrt{x^2+4}+C$$

Answered by MJS last updated on 11/Apr/19

$$\int x^2\sqrt{x^2+4}dx=$$$$[t={\tan }^{-1}\frac{x}{2}\to dx={2\sec }^{2}tdt]$$$$=16\int ({\sec }^{5}t-{\sec }^{3}t)dt=$$$$[\int {\sec }^{n}tdt=\frac{{\sec }^{n-2}t\tan t}{t-1}+\frac{n-2}{n-1}\int {\sec }^{n-2}tdt]$$$$=-2\ln (\tan t+\sec t)+\sec t(1-2\tan t(1-{2\sec }^{2}t))=$$$$=\frac{1}{4}x(x^2+2)\sqrt{x^2+4}-2\ln \mid x+\sqrt{x^2+4}\mid +C$$

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

$$thankssirmjs$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com