What is the integral from 0 to a of sin(cx)sin(cx) ?

The integral from 0 to a of sin(cx)sin(cx) is 1/2 a-1/(4c)sin(2ca)

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integral from 0 to a of sin(cx)sin(cx)

\int_{0}^{a} sin (c x) sin (c x) d x

\frac{1}{2} a - \frac{1}{4 c} sin (2 c a)

Solution steps

\int_{0}^{a} sin (c x) sin (c x) d x

Simplify

sin (c x) sin (c x) : sin^{2} (c x)

= \int_{0}^{a} sin^{2} (c x) d x

Rewrite using trig identities

= \int_{0}^{a} 1 - cos^{2} (c 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}^{a} 1 d x - \int_{0}^{a} cos^{2} (c x) d x

\int_{0}^{a} 1 d x = a

\int_{0}^{a} cos^{2} (c x) d x = \frac{1}{2} (a + \frac{sin ( 2 c a )}{2 c})

= a - \frac{1}{2} (a + \frac{sin ( 2 c a )}{2 c})

Simplify

a - \frac{1}{2} (a + \frac{sin ( 2 c a )}{2 c}) : \frac{1}{2} a - \frac{1}{4 c} sin (2 c a)

= \frac{1}{2} a - \frac{1}{4 c} sin (2 c a)

\int_{1}^{e} \frac{4 ( 1 + ln ( x ) )}{x} d x

\int_{- 3}^{3} 9 d x

\int_{0}^{1} (2 - t^{2}) d t

\int_{2}^{e + 1} (\frac{x}{x - 1}) d x

\int_{0}^{x} sin (4 t^{2}) d t

What is the integral from 0 to a of sin(cx)sin(cx) ?
The integral from 0 to a of sin(cx)sin(cx) is 1/2 a-1/(4c)sin(2ca)