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I know that:$$\int \cos x dx = \sin x +C$$ Substiute $x$ for $ax+b$: $$\int \cos(ax+b) dx = \sin(ax+b) +C$$ but according to my book: $$\int \cos(ax+b) dx = \frac{1}{a}\sin(ax+b) +C$$ Why doesn't substiuting work here?

user773458
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BadUsername
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    If $y=ax+b$ then $dy=adx$. – John Douma Feb 20 '24 at 15:14
  • Think about the chain rule. Since $\frac{d}{dx}f(g(x))=f'(g(x))g'(x)$, $\int f'(g(x))g'(x)dx=f(g(x))+C=\int f'(y)dy$ with the substitution $y=g(x)$. – J.G. Feb 20 '24 at 15:21
  • @JohnDouma where can I learn more about what you just said? – BadUsername Feb 20 '24 at 17:02
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    @BadUsername You have to convert the differential when you apply substitution. This should be in the examples in your calculus book. – John Douma Feb 20 '24 at 17:21
  • If your book has just gotten to $\int\cos xdx=\sin x+C$, then you'll want to look a chapter or so ahead for the section titled "$u$-substitution". – Teepeemm Feb 21 '24 at 02:21

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You need to substitute $x$ for $ax + b$ not only in argument of $\cos$, but also in differential: $\int \cos(ax + b)\, d\color{red}{(ax + b)} = \sin(ax + b)$.

This is equivalent to your citation, as $\int \cos(ax + b)\, d(ax + b) = a \cdot \int \cos(ax + b)\, dx$.

mihaild
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