Solution assignment 07 Integration by substitution

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Assignment 7

Solve:

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

Solution

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

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