Integral of $$$9 t \sin{\left(t \right)} \cos{\left(t \right)}$$$

Find $$$\int 9 t \sin{\left(t \right)} \cos{\left(t \right)}\, dt$$$.

Apply the constant multiple rule $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ with $$$c=9$$$ and $$$f{\left(t \right)} = t \sin{\left(t \right)} \cos{\left(t \right)}$$$:

$${\color{red}{\int{9 t \sin{\left(t \right)} \cos{\left(t \right)} d t}}} = {\color{red}{\left(9 \int{t \sin{\left(t \right)} \cos{\left(t \right)} d t}\right)}}$$

For the integral $$$\int{t \sin{\left(t \right)} \cos{\left(t \right)} d t}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=t$$$ and $$$\operatorname{dv}=\sin{\left(t \right)} \cos{\left(t \right)} dt$$$.

Then $$$\operatorname{du}=\left(t\right)^{\prime }dt=1 dt$$$ (steps can be seen ») and $$$\operatorname{v}=\int{\sin{\left(t \right)} \cos{\left(t \right)} d t}=\frac{\sin^{2}{\left(t \right)}}{2}$$$ (steps can be seen »).

Therefore,

$$9 {\color{red}{\int{t \sin{\left(t \right)} \cos{\left(t \right)} d t}}}=9 {\color{red}{\left(t \cdot \frac{\sin^{2}{\left(t \right)}}{2}-\int{\frac{\sin^{2}{\left(t \right)}}{2} \cdot 1 d t}\right)}}=9 {\color{red}{\left(\frac{t \sin^{2}{\left(t \right)}}{2} - \int{\frac{\sin^{2}{\left(t \right)}}{2} d t}\right)}}$$

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

$$\frac{9 t \sin^{2}{\left(t \right)}}{2} - 9 {\color{red}{\int{\frac{\sin^{2}{\left(t \right)}}{2} d t}}} = \frac{9 t \sin^{2}{\left(t \right)}}{2} - 9 {\color{red}{\left(\frac{\int{\sin^{2}{\left(t \right)} d t}}{2}\right)}}$$

Apply the power reducing formula $$$\sin^{2}{\left(\alpha \right)} = \frac{1}{2} - \frac{\cos{\left(2 \alpha \right)}}{2}$$$ with $$$\alpha=t$$$:

$$\frac{9 t \sin^{2}{\left(t \right)}}{2} - \frac{9 {\color{red}{\int{\sin^{2}{\left(t \right)} d t}}}}{2} = \frac{9 t \sin^{2}{\left(t \right)}}{2} - \frac{9 {\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 t \right)}}{2}\right)d t}}}}{2}$$

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

$$\frac{9 t \sin^{2}{\left(t \right)}}{2} - \frac{9 {\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 t \right)}}{2}\right)d t}}}}{2} = \frac{9 t \sin^{2}{\left(t \right)}}{2} - \frac{9 {\color{red}{\left(\frac{\int{\left(1 - \cos{\left(2 t \right)}\right)d t}}{2}\right)}}}{2}$$

Integrate term by term:

$$\frac{9 t \sin^{2}{\left(t \right)}}{2} - \frac{9 {\color{red}{\int{\left(1 - \cos{\left(2 t \right)}\right)d t}}}}{4} = \frac{9 t \sin^{2}{\left(t \right)}}{2} - \frac{9 {\color{red}{\left(\int{1 d t} - \int{\cos{\left(2 t \right)} d t}\right)}}}{4}$$

Apply the constant rule $$$\int c\, dt = c t$$$ with $$$c=1$$$:

$$\frac{9 t \sin^{2}{\left(t \right)}}{2} + \frac{9 \int{\cos{\left(2 t \right)} d t}}{4} - \frac{9 {\color{red}{\int{1 d t}}}}{4} = \frac{9 t \sin^{2}{\left(t \right)}}{2} + \frac{9 \int{\cos{\left(2 t \right)} d t}}{4} - \frac{9 {\color{red}{t}}}{4}$$

Let $$$u=2 t$$$.

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

The integral can be rewritten as

$$\frac{9 t \sin^{2}{\left(t \right)}}{2} - \frac{9 t}{4} + \frac{9 {\color{red}{\int{\cos{\left(2 t \right)} d t}}}}{4} = \frac{9 t \sin^{2}{\left(t \right)}}{2} - \frac{9 t}{4} + \frac{9 {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{4}$$

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 $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

$$\frac{9 t \sin^{2}{\left(t \right)}}{2} - \frac{9 t}{4} + \frac{9 {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{4} = \frac{9 t \sin^{2}{\left(t \right)}}{2} - \frac{9 t}{4} + \frac{9 {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}}{4}$$

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

$$\frac{9 t \sin^{2}{\left(t \right)}}{2} - \frac{9 t}{4} + \frac{9 {\color{red}{\int{\cos{\left(u \right)} d u}}}}{8} = \frac{9 t \sin^{2}{\left(t \right)}}{2} - \frac{9 t}{4} + \frac{9 {\color{red}{\sin{\left(u \right)}}}}{8}$$

Recall that $$$u=2 t$$$:

$$\frac{9 t \sin^{2}{\left(t \right)}}{2} - \frac{9 t}{4} + \frac{9 \sin{\left({\color{red}{u}} \right)}}{8} = \frac{9 t \sin^{2}{\left(t \right)}}{2} - \frac{9 t}{4} + \frac{9 \sin{\left({\color{red}{\left(2 t\right)}} \right)}}{8}$$

Therefore,

$$\int{9 t \sin{\left(t \right)} \cos{\left(t \right)} d t} = \frac{9 t \sin^{2}{\left(t \right)}}{2} - \frac{9 t}{4} + \frac{9 \sin{\left(2 t \right)}}{8}$$

Simplify:

$$\int{9 t \sin{\left(t \right)} \cos{\left(t \right)} d t} = \frac{9 \left(4 t \sin^{2}{\left(t \right)} - 2 t + \sin{\left(2 t \right)}\right)}{8}$$

Add the constant of integration:

$$\int{9 t \sin{\left(t \right)} \cos{\left(t \right)} d t} = \frac{9 \left(4 t \sin^{2}{\left(t \right)} - 2 t + \sin{\left(2 t \right)}\right)}{8}+C$$

$$$\int 9 t \sin{\left(t \right)} \cos{\left(t \right)}\, dt = \frac{9 \left(4 t \sin^{2}{\left(t \right)} - 2 t + \sin{\left(2 t \right)}\right)}{8} + C$$$A