Question

Solve each of the definite integrals in Exercises 67–76.

${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\cos }^3\left(x\right)dx$

Short answer

The solution of the integral is$\frac{4}{3}$.

Step-by-step solution

Step 1. Given Information

The given integral is${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\cos }^3\left(x\right)dx$

Step 2. Rewrite and substitute

• Use the trigonometric identities to rewrite the given integral.

${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\cos }^3\left(x\right)dx={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\left(1-{\sin }^2\left(x\right)\right)\cos \left(x\right)dx$

• Assume that $\sin \left(x\right)=u$. So, $\cos \left(x\right)dx=du$.
• Substitute into the integral and change the limits accordingly.

$\begin{array}{rcl}{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\left(1-{\sin }^2\left(x\right)\right)\cos \left(x\right)dx& =& {\int }_{{\sin }^{-1}\left(-\frac{\pi }{2}\right)}^{{\sin }^{-1}\left(\frac{\pi }{2}\right)}1-u^{2}du\\ & =& {\int }_{-1}^{1}1-u^{2}du\end{array}$

Step 3. Integrate

• Integrate the obtained integral and substitute the limits to find the solution.

$\begin{array}{rcl}{\int }_{-1}^{1}1-u^{2}du& =& {\int }_{-1}^{1}1du-{\int }_{-1}^1{u}^{2}du\\ & =& {\left[u-\frac{u^3}{3}\right]}_{-1}^1\\ & =& 1-\frac{1}{3}+1-\frac{1}{3}\\ & =& \frac{4}{3}\end{array}$