Integral of $$$e^{9} \sin{\left(7 x \right)} \cos{\left(7 x \right)}$$$
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Your Input
Find $$$\int e^{9} \sin{\left(7 x \right)} \cos{\left(7 x \right)}\, dx$$$.
Solution
Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=e^{9}$$$ and $$$f{\left(x \right)} = \sin{\left(7 x \right)} \cos{\left(7 x \right)}$$$:
$${\color{red}{\int{e^{9} \sin{\left(7 x \right)} \cos{\left(7 x \right)} d x}}} = {\color{red}{e^{9} \int{\sin{\left(7 x \right)} \cos{\left(7 x \right)} d x}}}$$
Let $$$u=\sin{\left(7 x \right)}$$$.
Then $$$du=\left(\sin{\left(7 x \right)}\right)^{\prime }dx = 7 \cos{\left(7 x \right)} dx$$$ (steps can be seen »), and we have that $$$\cos{\left(7 x \right)} dx = \frac{du}{7}$$$.
Therefore,
$$e^{9} {\color{red}{\int{\sin{\left(7 x \right)} \cos{\left(7 x \right)} d x}}} = e^{9} {\color{red}{\int{\frac{u}{7} d u}}}$$
Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{7}$$$ and $$$f{\left(u \right)} = u$$$:
$$e^{9} {\color{red}{\int{\frac{u}{7} d u}}} = e^{9} {\color{red}{\left(\frac{\int{u d u}}{7}\right)}}$$
Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=1$$$:
$$\frac{e^{9} {\color{red}{\int{u d u}}}}{7}=\frac{e^{9} {\color{red}{\frac{u^{1 + 1}}{1 + 1}}}}{7}=\frac{e^{9} {\color{red}{\left(\frac{u^{2}}{2}\right)}}}{7}$$
Recall that $$$u=\sin{\left(7 x \right)}$$$:
$$\frac{e^{9} {\color{red}{u}}^{2}}{14} = \frac{e^{9} {\color{red}{\sin{\left(7 x \right)}}}^{2}}{14}$$
Therefore,
$$\int{e^{9} \sin{\left(7 x \right)} \cos{\left(7 x \right)} d x} = \frac{e^{9} \sin^{2}{\left(7 x \right)}}{14}$$
Add the constant of integration:
$$\int{e^{9} \sin{\left(7 x \right)} \cos{\left(7 x \right)} d x} = \frac{e^{9} \sin^{2}{\left(7 x \right)}}{14}+C$$
Answer
$$$\int e^{9} \sin{\left(7 x \right)} \cos{\left(7 x \right)}\, dx = \frac{e^{9} \sin^{2}{\left(7 x \right)}}{14} + C$$$A