How to solve π3π3 (|x|+tanx)2dx?

x is an absolute value

1 Answer
Jun 8, 2018

π3π3(|x|+tanx)2dx=2π381π3+3

Explanation:

Using the additivity of the integral:

π3π3(|x|+tanx)2dx=0π3(|x|+tanx)2dx+π30(|x|+tanx)2dx

In the first integral substitute t=x:

0π3(|x|+tanx)2dx=0π3(|t|+tan(t))2d(t)

0π3(|x|+tanx)2dx=π30(|t|tant)2dt

Change the name of the integration variable to x again for clarity:

π3π3(|x|+tanx)2dx=π30(|x|+tanx)2dx+π30(|x|tanx)2dx

as in the interval of integration x is positive, then |x|=x:

π3π3(|x|+tanx)2dx=π30(x+tanx)2dx+π30(xtanx)2dx

using the linearity of the integral:

π3π3(|x|+tanx)2dx=π30((x+tanx)2+(xtanx)2)dx

π3π3(|x|+tanx)2dx=π30(x2+2xtanx+tan2x+x22xtanx+tan2x)dx

π3π3(|x|+tanx)2dx=2π30(x2+tan2x)dx

π3π3(|x|+tanx)2dx=2π30x2dx+2π30tan2xdx

Using the trigonometric identity:

tan2x=sec2x1

π3π3(|x|+tanx)2dx=2π30x2dx+2π30(sec2x1)dx

π3π3(|x|+tanx)2dx=2π30x2dx+2π30sec2xdx2π30dx

π3π3(|x|+tanx)2dx=2[x33+tanxx]π30

π3π3(|x|+tanx)2dx=2π381π3+3