How do you integrate #int xsqrt(1-x^4)# by trigonometric substitution?

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Answer 1 · 2016-09-06T04:35:29

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

We have:

#I=intxsqrt(1-x^4)dx#

Let #x^2=sin(u)#. This implies that #2xdx=cos(u)du#. So, we have:

#I=1/2int2xsqrt(1-(x^2)^2)dx#

#I=1/2intcos(u)sqrt(1-sin^2(u))du#

Since #sin^2(u)+cos^2(u)=1#, we see that #cos(u)=sqrt(1-sin^2(u)#:

#I=1/2intcos^2(u)du#

To integrate this, recall that #cos(2u)=2cos^2(u)-1#, so #cos^2(u)=1/2(cos(2u)-1)#. Plugging this into the integral:

#I=1/4int(cos(2u)-1)du#

#I=1/4intcos(2u)du-1/4intdu#

Solving both of these integrals:

#I=1/8int2cos(2u)du-1/4u#

#I=1/8sin(2u)-1/4u#

#I=1/8(2sin(u)cos(u))-1/4u#

#I=1/4sin(u)cos(u)-1/4u#

Write the cosine function as sine, since we are working with a sine-based substitution:

#I=1/4sin(u)sqrt(1-sin^2(u))-1/4u#

Plugging in #sin(u)=x^2# and #u=arcsin(x^2)#:

#I=1/4(x^2)sqrt(1-(x^2)^2)-1/4arcsin(x^2)#

#I=(x^2sqrt(1-x^4)-arcsin(x^2))/4+C#