Tip: Are you a non-native English speaker? I have just finished creating a 
Web AppIntegrating the third power of \sin(x) (or any odd power, for that matter), is an easy task (unlike ∫ \sin^2(x)\,dx, which requires a little trick). All you have to do is write the expression as \sin(x)⋅(\text{even power of }\sin), rewrite the even power using the formula \sin^2(x) = 1-\cos^2(x), and apply the substitution u = \cos(x) (i.e. du = -\sin(x)dx). Let’s see it in practice:
∫ \sin^3(x)\,dx = ∫ \sin(x)\sin^2(x)\,dx = ∫ \sin(x)(1-\cos^2(x))\,dx = \l.∫ -(1-u^2)\,du\r|_{u = \cos(x)} = -u + \frac{u^3}3 + c = \frac13 \cos^3(x)-\cos(x) + c\,.The same approach works for any odd power of \sin(x) (or \cos(x)), only the resulting expressions get slightly more complicated. For example,
∫ \sin^5(x)\,dx = ∫ \sin(x)(\sin^2(x))^2\,dx = ∫ \sin(x)(1-\cos^2(x))^2\,dx \\ = ∫ -(1-u^2)^2\,du = ∫ (-u^4+2u^2-1)\,du = -\frac{u^5}5 + \frac{2u^3}3 - u + c \\ = -\frac15 \cos^5(x) + \frac23 \cos^3(x)-\cos(x) + c\,.
