Trig Integral Example - # 3

[Jason76]

I made this one up so to see the way to solve it. I'm thinking a problem like this might not come into existence, and it isn't solvable.

\(\displaystyle \int \cos^{2}9x \sin x dx\)

\(\displaystyle u = \cos 9x\)

\(\displaystyle du = -\sin 9x dx\)

\(\displaystyle -\dfrac{1}{9}du = \sin x dx\)

\(\displaystyle -\dfrac{1}{9} \int u^{2} du\)

\(\displaystyle \rightarrow -\dfrac{1}{9} \dfrac{u^{3}}{3} + C\)

\(\displaystyle \rightarrow -\dfrac{u^{3}}{27} + C\)

\(\displaystyle \rightarrow -\dfrac{\sin^{3}9x}{27} + C\)

 

[pka]

I made this one up so to see the way to solve it. I'm thinking a problem like this might not come into existence, and it isn't solvable.

\(\displaystyle \int \cos^{2}9x \sin x dx\)

\(\displaystyle u = \cos 9x\)

\(\displaystyle du = -\sin 9x dx\)

\(\displaystyle -\dfrac{1}{9}du = \sin x dx\)

\(\displaystyle -\dfrac{1}{9} \int u^{2} du\)

\(\displaystyle \rightarrow -\dfrac{1}{9} \dfrac{u^{3}}{3} + C\)

\(\displaystyle \rightarrow -\dfrac{u^{3}}{27} + C\)

\(\displaystyle \rightarrow -\dfrac{\sin^{3}9x}{27} + C\)

Your answer is incorrect

\(\displaystyle \int \cos^{2}9x \sin (9x) dx=\dfrac{-\cos^3(9x)}{27}+C\)

Study the correct answer discover your mistakes.

 

[srmichael]

Jason, did you really mean

\(\displaystyle \int \cos^{2}(9x) \sin(x) dx\)

or

\(\displaystyle \int \cos^{2}(9x) \sin(9x) dx\)

Because the second integral leads to the answer pka provided. I think pka did not notice that the sine function was sin(x) and not sin(9x).

 

[HallsofIvy]

IF it really was \(\displaystyle \int cos^2(9x) sin(x) dx\) then you are probably best using
\(\displaystyle cos(9x)= \dfrac{e^{9ix}+ e^{-9ix}}{2}\) and \(\displaystyle sin(x)= \dfrac{e^{ix}- e^{-ix}}{2i}\)