Class 10
\[\int \cos^m x \sin^n x\;dx\]
If \(m\) is odd, split \(\cos^m x\) as \(\cos^{m-1} x\cos x\) and substitute \(u = \sin x\).
If \(n\) is odd, split \(\sin^m x\) as \(\sin^{m-1} x\sin x\) and substitute \(u = \cos x\).
If both \(m\) and \(n\) are even, use power reduction formulas.
\[\begin{aligned} \int_0^{\pi/2} \sin^3 x\cos^{11}x\;dx &\class{fragment}{{}=\int_0^{\pi/2} \sin^3 x\cos^{10}x\cos x\;dx} \\\\ &\class{fragment}{{}=\int_0^{\pi/2} \sin^3 x\left(\cos^{2}x\right)^5\cos x\;dx} \\\\ &\class{fragment}{{}=\int_0^{\pi/2} \sin^3 x\left(1-\sin^{2}x\right)^5\cos x\;dx} \\\\ &\class{fragment}{\color{blue}u = \sin x, \qquad du = \cos x\;dx} \\\\ &\class{fragment}{{}=\int_0^1 u^3\left(1-u^{2}\right)^5\;du} \\\\ &\class{fragment}{{}=\int_0^1 u^2\left(1-u^{2}\right)^5u\;du} \\\\ &\class{fragment}{\color{green}t = 1 - u^2, \qquad dt = -2u\;du, \qquad u^2 = 1 - t} \\\\ &\class{fragment}{{}=-\frac{1}{2}\int_1^0 (1-t)t^5\;dt} \\\\ &\class{fragment}{{}=\frac{1}{2}\int_0^1 t^5 - t^6\;dt} \\\\ &\class{fragment}{{}=\frac{1}{2}\left.\left(\frac{t^6}{6} - \frac{t^7}{7}\right)\right\rvert_0^1 = \frac{1}{12} - \frac{1}{14}} \end{aligned}\]
\[\begin{aligned} \int_0^{\pi/2} \sin^2 x\cos^{6}x\;dx &\class{fragment}{{}=\int_0^{\pi/2} \sin^2 x\left(\cos^{2}x\right)^3\;dx} \\\\ &\class{fragment}{{}=\int_0^{\pi/2} \frac{1 - \cos(2x)}{2} \left(\frac{1+\cos(2x)}{2}\right)^3\;dx} \\\\ &\class{fragment}{{}=\int_0^{\pi/2} \frac{1 - \cos(2x)}{2} \frac{1+\cos(2x)}{2}\left(\frac{1+\cos(2x)}{2}\right)^2\;dx} \\\\ &\class{fragment}{{}=\int_0^{\pi/2} \frac{1 - \cos^2(2x)}{4}\frac{1+2\cos(2x)+\cos^2(2x)}{4}\;dx} \\\\ &\class{fragment}{{}=\frac{1}{16}\int_0^{\pi/2} \sin^2(2x)\left(1+2\cos(2x)+\cos^2(2x)\right)\;dx} \end{aligned}\]
\[\int_0^{\pi/2} \sin^2(2x)\left(1+2\cos(2x)+\cos^2(2x)\right)\;dx\]