What is the integral from 0 to pi of sin(2x)-sin(x) ?

The integral from 0 to pi of sin(2x)-sin(x) is -2

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integral from 0 to pi of sin(2x)-sin(x)

\int_{0}^{π} sin (2 x) - sin (x) d x

- 2

Solution steps

\int_{0}^{π} sin (2 x) - sin (x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{π} sin (2 x) d x - \int_{0}^{π} sin (x) d x

\int_{0}^{π} sin (2 x) d x = 0

\int_{0}^{π} sin (x) d x = 2

= 0 - 2

Simplify

= - 2

\int_{0}^{π} x sin (n) x d x

\int_{1}^{3} \frac{x ^{2}}{2} d x

\int_{0}^{π} sin^{2} (x) sin (n x) d x

\int_{0}^{t} sin (t - x) d x

\int_{0}^{1} x^{2} ln (1 + x^{2}) d x

What is the integral from 0 to pi of sin(2x)-sin(x) ?
The integral from 0 to pi of sin(2x)-sin(x) is -2