Home > Calculus > Integration by Trigonometric Substitution

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

Answer 1

Cameron Alexander

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#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Isabella Ackerman

To integrate ∫x/sqrt(16 - 9x^4) by trigonometric substitution, you can let x = (4/3)sin(θ). Then dx = (4/3)cos(θ)dθ. Substitute these expressions into the integral and simplify. You'll end up with an integral involving trigonometric functions that can be evaluated using trigonometric identities and basic integration techniques.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.