Evaluate #int(xsqrt(a^2-x^2))/sqrt(a^2+x^2) dx#?

Question

514 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-08T12:31:49

#a^2/2 sin^-1(x^2/a^2)+1/2sqrt(a^4-x^4)+C#

#int x sqrt{a^(2)-x^(2)}/sqrt{a^(2)+x^(2)} dx = int ((a^2-x^2)xdx)/sqrt(a^4-x^4)#

Substitute #x^2 = a^2 sin theta# so that we get

#2xdx = a^2 cos theta d theta, qquad sqrt(a^4-x^4) = a^2 cos theta#

Thus the integral becomes

# int (1/2a^2(1-sin theta)a^2 cos theta d theta)/(a^2 cos theta) = a^2/2int (1-sin theta) d theta#

# qquad = a^2/2 (theta + cos theta)+C#

#qquad = a^2/2 sin^-1(x^2/a^2)+1/2sqrt(a^4-x^4)+C#