How do you find#int (x^2) / (sqrt(81 - x^2)) dx # using trigonometric substitution?

1 Answer
Trevor Ryan.
Dec 6, 2015

#=81/2sin^(-1)(x/9)-(xsqrt(81-x^2))/2+C#

Explanation:

Let #x=9sintheta#
Then #dx=9costheta d theta# and #sintheta=x/9#

Then the original integral becomes

#9int(81sin^2theta)/(sqrt(81-81sin^2theta))costheta d theta#

Now using the trig identities #sin^2theta+cos^2theta = 1# and
#sin2theta=2sinthetacostheta#, together with factorizations and laws of surds, we may compute the integral as

#9xx81int(sin^2theta)/(9sqrt(1-sin^2theta))costheta d theta#

#=81int(sin^2theta)/costheta costheta d theta#

#=81[theta/2-(sin2theta)/4]+C#

#=81/2sin^(-1)(x/9)-81/2*x/9*sqrt(81-x^2)/9+C#

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