How do you integrate #int x sqrt( 1 - x^4 )dx# using trigonometric substitution?

1 Answer
Mar 12, 2018

If we let #u = x^2#, then #du = 2xdx# and #dx= (du)/(2x)#.

#I = int xsqrt(1 - u^2)(du)/(2x)#

#I = 1/2intsqrt( 1 - u^2) du#

This is when the trig substitution comes in. Let #u = sintheta#. Then #du = costheta d theta#.

#I = 1/2int sqrt(1 - sin^2theta) costheta d theta#

#I = 1/2int sqrt(cos^2theta)costheta d theta#

#I = 1/2int costhetacostheta d theta#

#I = 1/2int cos^2theta d theta#

We know that #cos(2theta) = 2cos^2theta - 1#, so #1/2cos(2theta) = cos^2theta - 1/2# and #1/2cos(2theta) + 1/2 = cos^2theta#. This is the famous power reduction identity, frequently rewritten as #1/2(cos2theta + 1)#

#I = 1/2int 1/2(cos2theta + 1) d theta#

#I = 1/4int cos(2theta) + 1 d theta#

#I = 1/4(1/2sin(2theta)) + 1/4theta + C#

#I = 1/8sin(2theta) + 1/4theta + C#

Recall that #sin(2theta) = 2sinthetacostheta#.

#I = 1/8(2)sinthetacostheta + 1/4theta + C#

#I = 1/4sinthetacostheta + 1/4theta + C#

From our initial #theta#-substitution, we know that #sintheta = u/1#. Thus #theta = arcsin(u)# and #costheta = sqrt(1- u^2)#

#I = 1/4usqrt(1 - u^2) + 1/4arcsin(u) + C#

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

Hopefully this helps!