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Question Number 43832 by physics last updated on 16/Sep/18

Commented by physics last updated on 16/Sep/18

$$dowithfullexplanationplease$$

Commented by maxmathsup by imad last updated on 16/Sep/18

$$letA=\int \frac{dx}{{(x^2+a^2)}^{\frac{3}{2}}}changementx=ash\left(t\right)givet=argsh\left(\frac{x}{a}\right)and$$$$A=\int \frac{ach\left(t\right)dt}{a^3{\left({ch}^2\left(t\right)\right)}^{\frac{3}{2}}}=\frac{1}{a^2}\int \frac{dt}{{ch}^2\left(t\right)}(wesupposea\ne 0)$$$$A=\frac{2}{a^2}\int \frac{dt}{1+ch\left(2t\right)}=\frac{2}{a^2}\int \frac{dt}{1+\frac{e^{2t}+e^{-2t}}{2}}=\frac{4}{a^2}\int \frac{dt}{2+e^{2t}+e^{-2t}}$$$${=}_{e^{2t}=u}\frac{4}{a^2}\int \frac{1}{2+u+u^{-1}}\frac{du}{2u}=\frac{2}{a^2}\int \frac{du}{2u+u^2+1}=\frac{2}{a^2}\int \frac{du}{{(u+1)}^2}$$$$=\frac{-2}{a^2(u+1)}+c=-\frac{2}{a^2(1+e^{2t})}butt=ln(\frac{x}{a}+\sqrt{1+\frac{x^2}{a^2}})\Rightarrow$$$$e^{2t}={(\frac{x}{a}+\sqrt{1+\frac{x^2}{a^2}})}^2\Rightarrow A=\frac{-2}{a^2(1+{(\frac{x}{a}+\sqrt{1+\frac{x^2}{a^2}})}^2}+c$$$$ifa=0A=\int \frac{dx}{x^3}=\int x^{-3}dx=\frac{1}{-3+1}{x}^{-3+1}+c=\frac{-1}{2x^2}+c.$$

Answered by MJS last updated on 16/Sep/18

$$\int \frac{dx}{{(x^2+a^2)}^{\frac{3}{2}}}=$$$$[x=a\tan t\to t=\arctan \frac{x}{a}\to dx=a{\sec }^{2}tdt]$$$$=\int \frac{a{\sec }^{2}t}{{(a^2+a^2{\tan }^{2}t)}^{\frac{3}{2}}}dt=\int \frac{a{\sec }^{2}t}{a^3(1+{\tan }^{2}t)}dt=$$$$[1+{\tan }^{2}t=1+\frac{{\sin }^{2}t}{{\cos }^{2}t}=\frac{{\cos }^{2}t+{\sin }^{2}t}{\mathrm{c}o{\mathrm{s}}^{2}t}=\frac{1}{{\cos }^{2}t}={\sec }^{2}t]$$$$=\frac{1}{a^2}\int \frac{{\sec }^{2}t}{{\left({\sec }^{2}t\right)}^{\frac{3}{2}}}dt=\frac{1}{a^2}\int \frac{dt}{\sec t}=\frac{1}{a^2}\int \cos tdt=$$$$=\frac{1}{a^2}\sin t=\frac{1}{a^2}\sin \arctan \frac{x}{a}=$$$$[\sin \arctan t=\frac{t}{\sqrt{t^2+1}}]$$$$=\frac{x}{a^3\sqrt{\frac{x^2}{a^2}+1}}=\frac{x}{a^2\sqrt{x^2+a^2}}+C$$

Commented by physics last updated on 19/Sep/18

$$thankyousir$$

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