Chapter 5.5, Problem 49E

To determine

To evaluate: The definite integral.

Answer to Problem 49E

The integration of the definite integral ${\displaystyle \underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}\left(x^3+x^4\tan x\right)dx}$ is $0$ .

Explanation of Solution

Given information: ${\displaystyle \underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}\left(x^3+x^4\tan x\right)dx}$

Calculation: ${\displaystyle \underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}\left(x^3+x^4\tan x\right)dx}$

  $={\displaystyle \underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}{x}^{3}dx}+{\displaystyle \underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}{x}^4\tan xdx}\mathrm{......}(i)$

Assuming, $f\left(x\right)=x^3$

Therefore, $f\left(-x\right)={\left(-x\right)}^3=-x^3=-f\left(x\right)$

Hence, $f\left(-x\right)=-f\left(x\right)$

This is an odd function and continuous at $\left[-\frac{\pi }{4},\frac{\pi }{4}\right]$

Now, ${\displaystyle \underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}{x}^{3}dx}=\mathrm{0......}(ii)$

Again assume $g\left(x\right)=x^4\tan x$

Therefore, $g\left(-x\right)={\left(-x\right)}^4\tan \left(-x\right)=-x^4\tan x$

  $g\left(x\right)$ is also an odd function and continuous at $\left[-\frac{\pi }{4},\frac{\pi }{4}\right]$

Now, ${\displaystyle \underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}{x}^4\tan xdx}=\mathrm{0......}(iii)$

Putting the values of equation (ii) and (iii) in equation (i) we get,

  $\begin{array}{l}=0+0\\ =0\end{array}$

Hence, The integration of the definite integral ${\displaystyle \underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}\left(x^3+x^4\tan x\right)dx}$ is $0$ .