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What is the integral of #int sin^3 3x cos 3x dx#?

Answer 1

#intsin^3(3x)cos(3x)dx=frac{sin^4(3x)}{12}+c#

Take the given equation #intsin^3(3x)cos3xdx# Take the function #sin(3x)=t# Now differentiate with respect to #t# on both sides of this #\frac{d}{dt}(sin3x)=1\implies3cos3x*\frac{dx}{dt}=1\impliescos3xdx=dt/3#

Substituting the given above values for the main equation, we get #intt^3/3dt# This looks easy. Let's directly integrate. #intt^3/3dt=t^4/(3*4)+c=t^4/12+c#

That's what you want, right? Oh, substitution? That's there up in the answer section.

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Answer 2

Paisley Johnson

To find the integral of (\int \sin^3(3x) \cos(3x) , dx), you can use the trigonometric identity ( \sin^2(x) = 1 - \cos^2(x) ) to rewrite ( \sin^3(x) ) as ( \sin^2(x) \cdot \sin(x) ). Then apply the substitution (u = \sin(3x)) to simplify the integral. After integration, revert back to the original variable (x).

The result of the integral is: [ \frac{-\cos^4(3x)}{12} + C ] where ( C ) is the constant of integration.

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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.