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

The integral from 0 to a of (bsin(cx)+ is (b(-cos(ca)+1))/c+da

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integral from 0 to a of (bsin(cx)+

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

\frac{b ( - cos ( c a ) + 1 )}{c} + d a

Solution steps

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

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

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

= \frac{b ( - cos ( c a ) + 1 )}{c} + d a

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What is the integral from 0 to a of (bsin(cx)+ ?
The integral from 0 to a of (bsin(cx)+ is (b(-cos(ca)+1))/c+da