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 November 19th, 2012, 09:33 AM #1 Senior Member   Joined: Sep 2011 Posts: 395 Thanks: 0 Integrating Trigonometric Functions Can anyone help me confirm if I've solved this correctly? Many thanks. Q. By writing $\sin^3x=\sin x\sin^2x=\sin x(1-cos^2x)$ & by then using the substitution $u=\cos x$, show that $\int\sin^3dx=\frac{\cos^3x}{3}-\cos x+c$. Attempt: $\int\sin^3x\,dx=\int\sin x(1-\cos^2x)dx$ Change of Variable: $u=\cos x\rightarrow du=-\sin x\,dx\rightarrow\sin x\,dx=-du$ Therefore, $\int\sin x(1-\cos^2x)dx=\int-(1-u^2)du=\int(u^2-1)du=\frac{\cos^2x}{3}-\cos x+c$
 November 19th, 2012, 09:36 AM #2 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: Integrating Trigonometric Functions Looks almost good to me, you need 3 as the exponent on cos
 November 19th, 2012, 10:18 AM #3 Senior Member   Joined: Sep 2011 Posts: 395 Thanks: 0 Re: Integrating Trigonometric Functions Great, thank you.

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