# Integral of $\sin(u)\mathrm du$?

If I search for a general formula on google for $\displaystyle \int \sin(u) \ \mathrm du$, where $u$ represents a function, I am presented with $-\cos(u)+C$.

But, shouldn't the answer be $\dfrac{-\cos(u)}{u'}+C$?

This applies for integrals of basic trigonometric functions such as $\sec(u)$, $\csc(u)$, $\tan(u)$, etc. They all seem to be missing the division of u'.

• Why do you think should there be $u'$? – DHMO Apr 18 '17 at 14:37
• @DHMO For example, the integral of sin(4x) is (1/4)(-cos(4x)) + C. If the u' is not included, the answer is just -cos(4x), which is incorrect. – Carlos Jr. Apr 18 '17 at 14:40
• what would $\displaystyle \int \sin(x)\ \mathrm dx$ be, in your opinion? – DHMO Apr 18 '17 at 14:41
• @DHMO -cos(x) + C – Carlos Jr. Apr 18 '17 at 14:42
• @CarlosJr. No. $\int\sin(4x)dx=-\frac{1}{4}\cos(4x)$ and $\int\sin(4x)d(4x)=-\cos(4x)$. – user281392 Apr 18 '17 at 14:47

It is true that $\int \sin(au)\,du = -\frac{\cos(u)}{a} + C$, but in general $\int \sin(f(u))\,du \neq -\frac{\cos(f(u))}{f'(u)}$; there's no such formula. You can remember the valid case by a variable substitution: $\int \sin(au)\,du = \{x = au;\ dx = a\,du\} = \int \sin(x) \frac{dx}{a} = \frac{-\cos(x) + C_1}{a} = -\frac{\cos(x)}{a} + C_2 = -\frac{\cos(au)}{a} + C_2$.

Suppose we are finding derivative of $u$ with respect to $x$. Then we can write it as $\frac{du}{dx}$ or $u'$.

And if you are finding derivative with respect to $u$ then answer is 1.

But in case of integration it is not true with the case you are asking for.

We have to integrate $\sin u$. It is simply $-\cos u + c$.

If we can do in other way.

Put $\sin u = t$

$\cos u du = dt$

Then we have,

$\int \frac t{\cos u} dt$

$= \int \frac t{\sqrt {1-t^2}}$

$= - \frac 12 \int \frac {-2t}{\sqrt {1-t^2}} + c$

$= - \frac 12 \frac{\sqrt{(1-t^2)}}{\frac 12} + c$

Putting value of t,

$= - cos u + c$