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

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

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,

$${\color{red}{\int{t \sin{\left(t^{2} \right)} \cos{\left(t^{2} \right)} d t}}} = {\color{red}{\int{\frac{\sin{\left(2 u \right)}}{4} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{4}$$$ and $$$f{\left(u \right)} = \sin{\left(2 u \right)}$$$:

$${\color{red}{\int{\frac{\sin{\left(2 u \right)}}{4} d u}}} = {\color{red}{\left(\frac{\int{\sin{\left(2 u \right)} d u}}{4}\right)}}$$

Let $$$v=2 u$$$.

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

Therefore,

$$\frac{{\color{red}{\int{\sin{\left(2 u \right)} d u}}}}{4} = \frac{{\color{red}{\int{\frac{\sin{\left(v \right)}}{2} d v}}}}{4}$$

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

$$\frac{{\color{red}{\int{\frac{\sin{\left(v \right)}}{2} d v}}}}{4} = \frac{{\color{red}{\left(\frac{\int{\sin{\left(v \right)} d v}}{2}\right)}}}{4}$$

The integral of the sine is $$$\int{\sin{\left(v \right)} d v} = - \cos{\left(v \right)}$$$:

$$\frac{{\color{red}{\int{\sin{\left(v \right)} d v}}}}{8} = \frac{{\color{red}{\left(- \cos{\left(v \right)}\right)}}}{8}$$

Recall that $$$v=2 u$$$:

$$- \frac{\cos{\left({\color{red}{v}} \right)}}{8} = - \frac{\cos{\left({\color{red}{\left(2 u\right)}} \right)}}{8}$$

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

$$- \frac{\cos{\left(2 {\color{red}{u}} \right)}}{8} = - \frac{\cos{\left(2 {\color{red}{t^{2}}} \right)}}{8}$$

Therefore,

$$\int{t \sin{\left(t^{2} \right)} \cos{\left(t^{2} \right)} d t} = - \frac{\cos{\left(2 t^{2} \right)}}{8}$$

Add the constant of integration:

$$\int{t \sin{\left(t^{2} \right)} \cos{\left(t^{2} \right)} d t} = - \frac{\cos{\left(2 t^{2} \right)}}{8}+C$$

$$$\int t \sin{\left(t^{2} \right)} \cos{\left(t^{2} \right)}\, dt = - \frac{\cos{\left(2 t^{2} \right)}}{8} + C$$$A