Question

Evaluate: $\int {\sin }^{2}x{\cos }^{5}xdx$

• A. $\frac{1}{2}{\sin }^{2}x-\frac{2}{5}{\sin }^{5}x+\frac{2}{7}{\sin }^{7}x+c$
• B. $\frac{1}{2}{\sin }^{3}x-\frac{1}{5}{\sin }^{5}x+\frac{1}{7}{\sin }^{7}x+c$
• C. $\frac{1}{3}{\sin }^{2}x-\frac{1}{5}{\sin }^{5}x+\frac{2}{7}{\sin }^{7}x+c$
• D. $\frac{1}{3}{\sin }^{3}x-\frac{2}{5}{\sin }^{5}x+\frac{1}{7}{\sin }^{7}x+c$

Answer 1

The correct option is D $\frac{1}{3}{\sin }^{3}x-\frac{2}{5}{\sin }^{5}x+\frac{1}{7}{\sin }^{7}x+c$

Let $I=\int {\sin }^{2}x{\cos }^{5}xdx=\int {\sin }^{2}x(1-{\sin }^{2}x)(1-{\sin }^{2}x)\cos xdx$

Put $\sin x=t\Rightarrow \cos xdx=dt$

$I=\int t^2(1-t^2)(1-t^2)dt=\int (t^2-2t^4+t^6)dt$

$=\frac{t^3}{3}-\frac{2}{5}{t}^5+\frac{t^7}{7}+c=\frac{1}{3}{\sin }^{3}x-\frac{2}{5}{\sin }^{5}x+\frac{1}{7}{\sin }^{7}x+c$