How do you integrate #int sqrt(1-x^2)# by trigonometric substitution?

1 Answer
Steve M
Nov 28, 2016

# int sqrt(1-x^2)dx = (sin^-1x + xsqrt(1-x^2))/2 + C#

Explanation:

Let # I = int sqrt(1-x^2)dx #

The way to approach these types of problems is try and make a comparison with a known trig identity. In the case we will use:

# sin^2A + cos^2A -=1 => cos^2A=1-sin^2A #

If you compare to the integrand, we make the following substitution:

Let #x=sintheta => 1-sin^2theta=cos^2theta#
# :. 1-x^2 = cos^2theta #
# :. sqrt(1-x^2) = costheta #

And #dx/(d theta) = costheta#

Substituting into the integral we have:
# I = int cos theta cos theta d theta # =
# :. I = int cos^2 theta d theta #

Now #cos2A-=cos^2A-sin^2A = 2cos^2A-1 #
# :. cos^2A=1/2( 1+cos2A)#

And so;
# I = int 1/2( 1+cos2theta) d theta #
# 2I = int (1+cos2theta) d theta #
# 2I = theta + (sin2theta)/2 + C'#

We now need to use the identity #sin2A-=2sinAcosA#
# 2I = theta + sinthetacostheta + C'#
# :. 2I = sin^-1x + xsqrt(1-x^2) + C'#
# :. I = (sin^-1x + xsqrt(1-x^2) + C')/2#
# :. I = (sin^-1x + xsqrt(1-x^2))/2 + C#

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