Integral of $$$- \sin^{2}{\left(u \right)}$$$

Find $$$\int \left(- \sin^{2}{\left(u \right)}\right)\, du$$$.

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

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

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

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

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 $$$f{\left(u \right)} = 1 - \cos{\left(2 u \right)}$$$:

$$- {\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 u \right)}}{2}\right)d u}}} = - {\color{red}{\left(\frac{\int{\left(1 - \cos{\left(2 u \right)}\right)d u}}{2}\right)}}$$

Integrate term by term:

$$- \frac{{\color{red}{\int{\left(1 - \cos{\left(2 u \right)}\right)d u}}}}{2} = - \frac{{\color{red}{\left(\int{1 d u} - \int{\cos{\left(2 u \right)} d u}\right)}}}{2}$$

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:

$$\frac{\int{\cos{\left(2 u \right)} d u}}{2} - \frac{{\color{red}{\int{1 d u}}}}{2} = \frac{\int{\cos{\left(2 u \right)} d u}}{2} - \frac{{\color{red}{u}}}{2}$$

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,

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

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

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

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

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

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

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

Therefore,

$$\int{\left(- \sin^{2}{\left(u \right)}\right)d u} = - \frac{u}{2} + \frac{\sin{\left(2 u \right)}}{4}$$

Add the constant of integration:

$$\int{\left(- \sin^{2}{\left(u \right)}\right)d u} = - \frac{u}{2} + \frac{\sin{\left(2 u \right)}}{4}+C$$

$$$\int \left(- \sin^{2}{\left(u \right)}\right)\, du = \left(- \frac{u}{2} + \frac{\sin{\left(2 u \right)}}{4}\right) + C$$$A