Primitive of Sine of a x + b/Proof 2

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Corollary to Primitive of Sine Function

$\ds \int \map \sin {a x + b} \rd x = -\frac {\map \cos {a x + b} } a + C$


Proof

Let $u = a x + b$.

Then:

$\dfrac {\d u} {\d x} = a$

Then:

\(\ds \int \map \sin {a x + b} \rd x\) \(=\) \(\ds \int \dfrac {\sin u} a \rd u\) Integration by Substitution
\(\ds \) \(=\) \(\ds -\dfrac {\cos u} a\) Primitive of $\sin u$
\(\ds \) \(=\) \(\ds -\frac {\map \cos {a x + b} } a + C\) substituting back for $u$

$\blacksquare$


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