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Powers of sinus and cosinus are integrated in the following way:

  1. Odd powers

    $$\int \sin^{2n+1}(x)\,\mathrm{d}x=\int \sin^{2n}(x)\sin x\,\mathrm{d}x=\int (1-\cos^2(x))^n\sin x\,\mathrm{d}x$$

    Expanding the binomial $(1-\cos^2(x))^n$ results in several integrals of the form

    $$\int \cos^m(x)\sin x\,\mathrm{d}x$$

    all of which are solvable with the change of variable $t=\cos x$, $\mathrm{d}t=-\sin x \,\mathrm{d}x$.

    The integral

    $$\int \cos^{2n+1}(x)\,\mathrm{d}x=\int \cos^{2n}(x)\cos x\,\mathrm{d}x=\int (1-\sin^2(x))^n\cos x\,\mathrm{d}x$$

    is solved in an analogous manner

  2. Even powers

    $$\int \sin^{2n}(x)\,\mathrm{d}x=\int (\sin^2(x))^n\,\mathrm{d}x=\int \left(\dfrac{1-\cos(2x)}{2}\right)^n\,\mathrm{d}x$$
    $$\int \cos^{2n}(x)\,\mathrm{d}x=\int (\cos^2(x))^n\,\mathrm{d}x=\int \left(\dfrac{1+\cos(2x)}{2}\right)^n\,\mathrm{d}x$$

    and after expanding the binomial, the integrand is expressed in terms of lower powers of $\cos$; this way we may proceed iterating until everything is integrated