Integrating Trigonometric Functions with Indefinite Integrals

[Windowmaker]

Homework Statement

integrate cos ^3(πx-1)

Homework Equations

1-sin^2(πx-1) = cos ^2 (πx-1)

The Attempt at a Solution

(1-sin^2(πx-1))cos(πx-1)

Let u = πx-1
1/ πdu = dx
1-sin^2(u )cos (u)

1/ π(X-1/3sin^3(u)

Basically I'm Confused on how to get past the trig identity part.

 

[Windowmaker]

π is Suppose to be pi i guess..

 

[pergradus]

you don't really need to do a substitution for the argument of the cosine function, just use the trig identity like you did, factoring out a cos(∏x-1) and then make the subsitution u = sin(∏x -1) and du = ∏cos(∏x-1)dx and you get:

∫cos(∏x-1)dx -1/∏∫u²du

 

[Windowmaker]

I see the answer in my book, but I'm not entirely sure there are no cosines in the answer.

 

[Windowmaker]

Wow, that was more simple than i thought. Thank you! Its amazing how much an actual person can be when compared to a Textbook!

 

[dextercioby]

Don't forget the integration constant each time you don't have limits on the integral sign. Also the 'dx' if you're integrating wrt 'x' or 'du', if you're integrating wrt 'u', etc.

So that

∫ cos³(πx-1) dx = 'something(x)' + C

 

[LawrenceC]

At the risk of stating the obvious, if the answer is not provided and you work an integral, take the derivative of your answer to check to see if you wind up with the integrand once again.