How do you find the definite integral of #x*sqrt(1-x^4) dx# from #[0, 1]#?

Question

Emily Ammerman · Accepted Answer

#int_0^1xsqrt(1-x^4)dx=pi/8#

First without the bounds:

#intxsqrt(1-x^4)dx#

Let #x^2=sintheta#. This implies that #2xdx=costhetad theta#. We can then modify the integral:

#=1/2intsqrt(1-(x^2)^2)(2xdx)=1/2intsqrt(1-sin^2theta)(costhetad theta)=1/2intcos^2thetad theta#

Now apply the identity #cos^2theta=1/2(1+cos2theta)#:

#=1/4int(1+cos2theta)d theta=1/4intd theta+1/8intcos2theta(2d theta)=1/4theta+1/8sin2theta#

Using the identities #sin2theta=2sinthetacostheta# and #costheta=sqrt(1-sin^2theta)# along with the original substitution #x=sintheta#, we can rewrite this in terms of #x#:

#=1/4sin^-1x+1/8(2sinthetacostheta)=1/4sin^-1x+1/4sinthetasqrt(1-sin^2theta)#

#=1/4sin^-1x+1/4xsqrt(1-x^2)" "(+C)#

So the integral with bounds is:

#int_0^1xsqrt(1-x^4)dx=[1/4sin^-1x+1/4xsqrt(1-x^2)]_0^1#

#=1/4sin^-1 1+0-(0+0)=1/4(pi/2)=pi/8#