Answer to Question #225472 in Calculus for sbuda

Question

determine the integral from pie/2 to 0 of function sin2x cosx dx

Answer 1

$$\displaystyle\int_{\pi/2}^{0}\sin 2x \cos x dx$$

$$\int\sin2x \cos x dx=\int2\sin x\cos x \cos xdx$$

$$u=\cos x, du=-\sin x dx$$

$$\int2\sin x\cos x \cos xdx=-\int 2u^2du=-\dfrac{2}{3}u^3+C$$

$$=-\dfrac{2}{3}\cos^3x+C$$

$$\displaystyle\int_{\pi/2}^{0}\sin 2x \cos x dx=\big[-\dfrac{2}{3}\cos^3x\big]\begin{matrix} 0 \\ \pi/2 \end{matrix}=-\dfrac{2}{3}+0$$

$$\displaystyle\int_{\pi/2}^{0}\sin 2x \cos x dx=-\dfrac{2}{3}$$