Integral of $$$t \cos{\left(t^{2} \right)}$$$

Find $$$\int t \cos{\left(t^{2} \right)}\, dt$$$.

Let $$$u=t^{2}$$$.

Then $$$du=\left(t^{2}\right)^{\prime }dt = 2 t dt$$$ (steps can be seen »), and we have that $$$t dt = \frac{du}{2}$$$.

So,

$${\color{red}{\int{t \cos{\left(t^{2} \right)} d t}}} = {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

$${\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}} = {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}$$

The integral of the cosine is $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:

$$\frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{2} = \frac{{\color{red}{\sin{\left(u \right)}}}}{2}$$

Recall that $$$u=t^{2}$$$:

$$\frac{\sin{\left({\color{red}{u}} \right)}}{2} = \frac{\sin{\left({\color{red}{t^{2}}} \right)}}{2}$$

Therefore,

$$\int{t \cos{\left(t^{2} \right)} d t} = \frac{\sin{\left(t^{2} \right)}}{2}$$

Add the constant of integration:

$$\int{t \cos{\left(t^{2} \right)} d t} = \frac{\sin{\left(t^{2} \right)}}{2}+C$$

$$$\int t \cos{\left(t^{2} \right)}\, dt = \frac{\sin{\left(t^{2} \right)}}{2} + C$$$A