How do you integrate #int x/sqrt(16-9x^4)# by trigonometric substitution?

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Answer 1 · 2018-03-06T07:33:17

The answer is #=1/6*arcsin(3/4x^2)+ C#

Let's perform some simplification

#16-9x^4=16(1-9/16x^4)=16(1-(3/4x^2)^2)#

Perform the substitution

#sintheta=3/4x^2#, #=>#, #costheta d theta=3/4*2xdx#

#xdx=2/3cos theta d theta#

#16(1-(3/4x^2)^2)=16(1-sin^2theta)=16cos^2theta#

Therefore,

#int(xdx)/(sqrt(16-9x^4))=2/3int(costheta d theta)/(sqrt(16cos^2theta))#

#=1/6int d theta#

#=1/6theta#

#=1/6*arcsin(3/4x^2)+ C#