## University Calculus: Early Transcendentals (3rd Edition)

$$\int\sqrt{\sin x}\cos^3xdx=\frac{2(\sqrt{\sin x})^3}{2}-\frac{2(\sqrt{\sin x})^7}{2}+C$$
$$A=\int\sqrt{\sin x}\cos^3xdx$$ $$A=\int\sqrt{\sin x}\cos x\cos^2xdx$$ We can rewrite $\cos^2x=1-\sin^2x$. $$A=\int\sqrt{\sin x}\cos x(1-\sin^2x)dx$$ We set $u=\sqrt{\sin x}$. That makes $u^4=\sin^2x$. Then $$du=\frac{(\sin x)'}{2\sqrt{\sin x}}dx=\frac{\cos x}{2u}dx$$ That means, $$\cos xdx=2udu$$ Therefore, $$A=\int u(1-u^4)2udu=\int(2u^2-2u^6)du$$ $$A=\frac{2u^3}{3}-\frac{2u^7}{7}+C$$ $$A=\frac{2(\sqrt{\sin x})^3}{3}-\frac{2(\sqrt{\sin x})^7}{7}+C$$