Does anyone know how to do the above integral? I'm stumped on where to even start.
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$\displaystyle \int -9 \cos^3 (x) \sin^2 (x) dx =$ $\displaystyle -9 \int \cos (x) (1 - \sin^2 (x)) \sin^2 (x) dx =$ $\displaystyle -9 \int \sin^2 (x) \cos (x) - \sin^4 (x) \cos (x) dx$ Can you do it from here?
thanks for the help. From there you would just do u-sub with sinx, if I'm not mistaken
Originally Posted by cdlegendary thanks for the help. From there you would just do u-sub with sinx, if I'm not mistaken Yes, that will work. Then you have $\displaystyle u = \sin (x)$ $\displaystyle du = \cos (x) dx$ $\displaystyle -9 \int \sin^2(x) \cos (x) - \sin^4(x) \cos(x) dx =$ $\displaystyle -9 \int (u^2 - u^4) du$
And then subbing back in after taking the integral would give you ((9sinx^5)/5)-(3sinx^3)+C thanks for the help
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