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Question Number 12098 by Nayon last updated on 13/Apr/17

$$Evaluate\int \sqrt{x^2-a^2}dx$$$$$$$$$$

Answered by sma3l2996 last updated on 13/Apr/17

$$A=\int \sqrt{x^2-a^2}dx=a\int \sqrt{{\left(\frac{x}{a}\right)}^2-1}dx$$$$letu=\frac{x}{a}\Rightarrow du=\frac{dx}{a}$$$$A=a^2\int \sqrt{u^2-1}du$$$$letu=cosh\left(t\right)\Rightarrow du=sinh\left(t\right)dt$$$$sinh\left(t\right)du={sinh}^{2}tdt\iff \sqrt{{cosh}^{2}t-1}du={sinh}^{2}tdt$$$$\sqrt{u^2-1}du={sinh}^{2}tdt$$$$A=a^2\int {sinh}^{2}tdt=a^2\int \frac{cosh\left(2t\right)-1}{2}dt$$$$A=\frac{a^2}{2}(\frac{sinh\left(2t\right)}{2}-t)+C$$$$A=\frac{a^2}{4}(sinh\left(2acosh\left(\frac{x}{a}\right)\right)-acosh\left(\frac{x}{a}\right))+C$$

Commented by Nayon last updated on 13/Apr/17

$$plsanswithoutusinghyperbolic$$$$trig...$$$$$$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 14/Apr/17

$$x=ait,dx=aidt,i=\sqrt{-1}$$$$I=\int \sqrt{-a^2{t}^2-a^2}aidt=-a^2\int \sqrt{t^2+1}dt=$$$$=-a^2[t\sqrt{t^2+1}-\int t\frac{2tdt}{2\sqrt{t^2+1}}]=$$$$=-a^2[t\sqrt{t^2+1}-\int \frac{(t^2+1)-1}{\sqrt{t^2+1}}dt]=$$$$=-a^2[t\sqrt{t^2+1}-\int \sqrt{t^2+1}dt+\int \frac{dt}{\sqrt{t^2+1}}]=$$$$-a^{2}t\sqrt{t^2+1}+a^2\int \sqrt{t^2+1}dt-\frac{a^2}{2}ln(t+\sqrt{t^2+1})$$$$2I=-a^2\frac{x}{ai}.\frac{\sqrt{x^2-a^2}}{ai}-\frac{a^2}{2}ln(\frac{x}{ai}+\frac{\sqrt{x^2-a^2}}{ai})\Rightarrow$$$$I=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{4}ln(x+\sqrt{x^2-a^2})+C.◼$$$$(-a^2\times \frac{1}{ai}\times \frac{1}{ai}=1,C=\frac{a^2}{4}ln\left(ai\right))$$

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