How do you integrate #int x /sqrt( 81 - x^4 )dx# using trigonometric substitution?

Question

Christopher Agresta · Accepted Answer

#1/2arcsin(x^2/9)+C# We should apply the substitution #x^2=9sintheta#. Note that this implies that #2xdx=9costhetad theta#. #intx/sqrt(81-x^4)dx=1/2int(2xdx)/sqrt(81-(x^2)^2)=1/2int(9costhetad theta)/sqrt(81-81sin^2theta)# Note that #9/sqrt81=1#: #=1/2int(costhetad theta)/sqrt(1-sin^2theta)# Recall that #1-sin^2theta=cos^2theta#: #=1/2intd theta=1/2theta+C# From #x^2=9sintheta# we see that #theta=arcsin(x^2/9)#: #=1/2arcsin(x^2/9)+C#

Paisley Johnson · Answer

undefined To integrate $ \int \frac{x}{\sqrt{81 - x^4}} \, dx $ using trigonometric substitution, you can substitute $ x = 3\sin(\theta) $. This substitution leads to $ dx = 3\cos(\theta) \, d\theta $. After substitution, the integral becomes $ \int \frac{3\sin(\theta)}{\sqrt{81 - (3\sin(\theta))^4}} \cdot 3\cos(\theta) \, d\theta $. Simplifying, this becomes $ \int \frac{9\sin(\theta)\cos(\theta)}{\sqrt{81 - 81\sin^4(\theta)}} \, d\theta $. Simplify further to $ \int \frac{9\sin(\theta)\cos(\theta)}{\sqrt{81\cos^4(\theta)}} \, d\theta $. Then $ \int \frac{9\sin(\theta)\cos(\theta)}{9\cos^2(\theta)} \, d\theta $. This simplifies to $ \int \sin(\theta) \, d\theta $. Integrate to get $ -\cos(\theta) + C $. Finally, substitute back $ x = 3\sin(\theta) $ to get $ -\sqrt{81 - x^4} + C $.