Use the substitution \(u=\sin x\) to show that \(\displaystyle\int\sin^nx\cos x\,\mathrm dx=\dfrac{\sin^{n+1}x}{n+1}+C\), where \(n\) is a positive integer.[2]
By writing \(\sin^3x=\sin x\left(1-\cos^2x\right)\), find \(\displaystyle\int\sin^3x\,\mathrm dx\).[3]
Explain why the method of (b) works for \(\displaystyle\int\sin^5x\,\mathrm dx\) but not for \(\displaystyle\int\sin^4x\,\mathrm dx\). Use the identity \(\sin^2x=\dfrac{1-\cos2x}{2}\) twice to find \(\displaystyle\int\sin^4x\,\mathrm dx\).[4]
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Answer:(a) shown (b) \(-\cos x+\dfrac{\cos^3x}{3}+C\) (c) an odd power leaves a factor \(\sin x\) for \(\mathrm du\), an even power does not; \(\dfrac{3x}{8}-\dfrac{\sin2x}{4}+\dfrac{\sin4x}{32}+C\)