Question

Solve the following integral.

$\int {\cos }^{3}x{\sin }^{4}xdx$

Short answer

Answer is$\frac{{\sin }^{5}x}{5}-\frac{{\sin }^{7}x}{7}+C$

Step-by-step solution

Step 1. Given information

An integral is$\int {\cos }^{3}x{\sin }^{4}xdx$

Step 2. Explanation

$\int {\cos }^{3}x{\sin }^{4}xdx=\int \left(1-{\sin }^{2}x\right)\cos x{\sin }^{4}xdx$

Let width="61">sinx=u

$\begin{array}{rcl}\int {\cos }^{3}x{\sin }^{4}xdx& =& \int u^4\left(1-u^2\right)du\\ & =& \int u^4-u^{6}du\\ & =& \frac{u^5}{5}-\frac{u^7}{7}+C\\ & =& \frac{{\sin }^{5}x}{5}-\frac{{\sin }^{7}x}{7}+C\end{array}$