Solution assignment 07 Integration by substitution

Solve:

$\displaystyle\int_{0}^{\pi}[2\cos^2(x)-1]dx$

This integration can be simplified by applying a rule from trigonometry:

$\cos(2x)=2\cos^2(x)-1$

and thus the integral can be rewritten:

$\displaystyle\int_{0}^{\pi}[2\cos^2(x)-1]dx=\displaystyle\int_{0}^{\pi}\cos(2x)dx$

The primitive of the integrand is:

$\displaystyle\frac{1}{2}\sin(2x)$

We can verify this result by differentiation. The chain rule has to be applied.

So we get:

$\displaystyle\int_{0}^{\pi}\cos(2x)dx=\displaystyle\frac{1}{2}[\sin(2x)]_{0}^{\pi}=\displaystyle\frac{1}{2}[\sin(2\pi)-\sin(0)]=0$