How do I do this: $\displaystyle \displaystyle\int (\sin 2x.\sin x) dx$ --------------------------------------------- The Answer is: $\displaystyle \frac {2}{3} \sin^3x + c $ Thank you advance.
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$\displaystyle \sin (2x)\sin (x)dx = 2\sin ^2 (x)\left[ {\cos (x)dx} \right]$
Originally Posted by Plato $\displaystyle \sin (2x)\sin (x)dx = 2\sin ^2 (x)\left[ {\cos (x)dx} \right]$ How is cosx cancelled out? When I use the cosx double angle rule to integrate sinx, I don't obtain the correct answer. Can you show the stages please?
No I will not do the problem for you. But what you have is $\displaystyle 2 u^2 du$.
Originally Posted by Air How is cosx cancelled out? When I use the cosx double angle rule to integrate sinx, I don't obtain the correct answer. Can you show the stages please? $\displaystyle sin(2x)=2sin(x)cos(x)$
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