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Establish formulas for the integrals of cos(a+bt) and sin(a+bt)

by RoRi

For real constants a,b with b \neq 0 prove the validity of the following formulas:

    \begin{align*} \int_0^x \cos (a+bt) \, dt &= \frac{1}{b} (\sin (a+bx) - \sin a), \\ \phantom{-} \int_0^x \sin (a+bt) \, dt &= -\frac{1}{b} (\cos (a + bx) - \cos a). \end{align*}

Proof. We compute as follows, using the formula for the cosine of a sum, and expansion/contraction of the interval of integration,

    \begin{align*} \int_0^x \cos (a+bt) \, dt &= \int_0^x (\cos a \cos (bt) - \sin a \sin (bt)) \, dt \\ &= \frac{\cos a}{b} \int_0^{bx} \cos t \, dt - \frac{\sin a}{b} \int_0^{bx} \sin t \, dt & (b \neq 0)\\ &= \frac{\cos a}{b} \sin (bx) - \frac{sin a}{b} (1-\cos (bx)) \\ &= \frac{1}{b} (\cos a \sin (bx) - \sin a + \sin a \cos (bx)) \\ &= \frac{1}{b} (\sin (a+bx) - \sin a). \end{align*}

And for the other formula we similarly compute,

    \begin{align*} \int_0^x \sin (a+bt) \, dt &= \int_0^x (\sin a \cos (bt) + \sin (bt) \cos a) \, dt \\ &= \frac{\sin a}{b} \int_0^{bx} \cos t \, dt - \frac{\cos a}{b} \int_0^{bx} \sin t \, dt & (b \neq 0)\\ &= \frac{\sin a}{b} \sin (bx) + \frac{\cos a}{b} (1 - \cos(bx)) \\ &= \frac{1}{b} (\sin a \sin (bx) + \cos a - \cos a \cos (bx)) \\ &= -\frac{1}{b} (\cos (a+bx) - \cos a). \qquad \blacksquare \end{align*}

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