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?

  • 4
    $\begingroup$ If $y=ax+b$ then $dy=adx$. $\endgroup$
    – John Douma
    Feb 20 at 15:14
  • $\begingroup$ 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 $y=g(x)$. $\endgroup$
    – J.G.
    Feb 20 at 15:21
  • $\begingroup$ @JohnDouma where can I learn more about what you just said? $\endgroup$ Feb 20 at 17:02
  • 1
    $\begingroup$ @BadUsername You have to convert the differential when you apply substitution. This should be in the examples in your calculus book. $\endgroup$
    – John Douma
    Feb 20 at 17:21
  • $\begingroup$ 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". $\endgroup$
    – Teepeemm
    Feb 21 at 2:21

1 Answer 1


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 $\int \cos(ax + b)\, d(ax + b) = a \cdot \int \cos(ax + b)\, dx$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .