Answer to Question #103217 in Calculus for Patrick Kariuki

Question #103217
Integrate x²dx/(9-x²)½
1
Expert's answer
2020-02-19T02:48:45-0500
x29x2dx\int\frac{x^2}{\sqrt{9-x^2}}dx

We will replace the variables:

x=3sint>dx=3costdtx=3\sin{t} -> dx=3\cos{t} dt.

27sin2tcost99sin2tdt=27sin2tcost31sin2tdt=9sin2tcostcostdt=9sin2tdt\int\frac{27\sin^2{t}\cos{t}}{\sqrt{9-9\sin^2{t}}}dt=\int \frac{27\sin^2{t}\cos{t}}{3\sqrt{1-\sin^2{t}}}dt=\int\frac{9\sin^2{t}\cos{t}}{\cos{t}}dt=9\int\sin^2{t} dt


Use transformation:

sin2t=1cos2t2\sin^2{t}=\frac{1-\cos{2t}}{2},

then we get

9sin2tdt=91cos2t2dt=92(1cos2t)dt=92(t12sin2t)+C=92(tsintcost)+C9\int \sin^2{t}dt=9\int \frac{1-\cos{2t}}{2}dt=\frac{9}{2}\int (1-\cos{2t})dt=\frac{9}{2}(t-\frac{1}{2}\sin{2t})+C=\frac{9}{2}(t-\sin{t}\cos{t})+C

where CC - contact.


t=arcsinx3t=\arcsin{\frac{x}{3}}

sint=x3\sin{t}=\frac{x}{3}

cost=1sin2t=1x29\cos{t}=\sqrt{1-\sin^2{t}}=\sqrt{1-\frac{x^2}{9}}


92(tsintcost)+C=92(arcsinx3x31x29)+C=9arcsinx3x9x22+C\frac{9}{2}(t-\sin{t}\cos{t})+C=\frac{9}{2}(\arcsin{\frac{x}{3}}-\frac{x}{3}\sqrt{1-\frac{x^2}{9}})+C=\frac{9\arcsin{\frac{x}{3}}-x\sqrt{9-x^2}}{2}+C


Answer:

x29x2dx=9arcsinx3x9x22+C\int\frac{x^2}{\sqrt{9-x^2}}dx=\frac{9\arcsin{\frac{x}{3}}-x\sqrt{9-x^2}}{2}+C


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