The integral of $\sin (x/2)$?

Question

From what I'm solving the integral of $\sin(x/2)$ is $-\cos(x/2)$ but, whenever I check the answer the correct integral is $-2\cos(x/2)$. I don't know where I'm wrong.

Please, help.

Answer 1

Just proceed by susbtitution:

Set $u=\frac x2$, so $\mathrm du=\frac12\mathrm d x\iff \mathrm d x=2\mathrm du$. So the integral becomes $$\int\sin\frac x2\,\color{red}{\mathrm d x}=\int\sin u\cdot \color{red}{2\mathrm du}=2\int\sin u\,\mathrm du=2(-\cos u)=-2\cos u=-2\cos\frac x2.$$

Answer 2

Use substitution to simply integrate that function $$I=\int \sin(x/2)\,dx$$ $$u=\frac x2 \implies du =\frac {dx}2 \implies dx=2\,du$$ $$I=\int \sin(x/2)\,dx=2\int \sin(u)\,du=-2\cos(u)=-2\cos(x/2).$$

Answer 3

In terms of integrating functions, think about what would happen if you differentiate the result of your integration.

You suppose $$\int f(x)\,dx=-\cos\left(\frac x2\right)$$ If so then: $$f(x)=\frac 12\sin\left(\frac x2\right)$$ whereas your $f(x)=\sin\left(\frac x2\right)$

The factor of $2$ that is missing is the reason the correct integral is $-2\cos\left(\frac x2\right)$

Answer 4

$$\int \sin\left(\frac x2\right)\,dx$$

Now apply $u$-substitution:$u=\frac x2$

Then differentiate with respect to $x$. we get $du=\frac{dx}{2}$

$dx=2du$ $$=\int \sin(u)(2)\,du$$ $$=2\int \sin(u)\,du$$ $$=-2\cos(u)$$ Substitute back $u=\frac x2$

Then we get, $$-2\cos\left(\frac x2\right)$$ $$\int \sin\left(\frac x2\right)\,dx=-2\cos\left(\frac x2\right)+C.$$

Answer 5

$$I=\int{\sin\frac{x}{2}}dx$$ let$$u=\frac{x}{2}$$ so$$\frac{du}{dx}=\frac{1}{2}$$ and $$dx=2du$$ so $$I=2\int{\sin(u)du}=-2\cos(u)+c=-2\cos\frac{x}{2}+c$$ If in doubt, use substitution.