Integral of 1/(-x*sin(1/2) + 1)

The calculator will find the integral/antiderivative of $$$\frac{1}{- x \sin{\left(\frac{1}{2} \right)} + 1}$$$, with steps shown.

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Find $$$\int \frac{1}{- x \sin{\left(\frac{1}{2} \right)} + 1}\, dx$$$.

The trigonometric functions expect the argument in radians. To enter the argument in degrees, multiply it by pi/180, e.g. write 45° as 45*pi/180, or use the appropriate function adding 'd', e.g. write sin(45°) as sind(45).

Solution

Let $$$u=- x \sin{\left(\frac{1}{2} \right)} + 1$$$.

Then $$$du=\left(- x \sin{\left(\frac{1}{2} \right)} + 1\right)^{\prime }dx = - \sin{\left(\frac{1}{2} \right)} dx$$$ (steps can be seen »), and we have that $$$dx = - \frac{du}{\sin{\left(\frac{1}{2} \right)}}$$$.

The integral can be rewritten as

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

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

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

The integral of $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

Recall that $$$u=- x \sin{\left(\frac{1}{2} \right)} + 1$$$:

$$- \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{\sin{\left(\frac{1}{2} \right)}} = - \frac{\ln{\left(\left|{{\color{red}{\left(- x \sin{\left(\frac{1}{2} \right)} + 1\right)}}}\right| \right)}}{\sin{\left(\frac{1}{2} \right)}}$$

Therefore,

$$\int{\frac{1}{- x \sin{\left(\frac{1}{2} \right)} + 1} d x} = - \frac{\ln{\left(\left|{x \sin{\left(\frac{1}{2} \right)} - 1}\right| \right)}}{\sin{\left(\frac{1}{2} \right)}}$$

Add the constant of integration:

$$\int{\frac{1}{- x \sin{\left(\frac{1}{2} \right)} + 1} d x} = - \frac{\ln{\left(\left|{x \sin{\left(\frac{1}{2} \right)} - 1}\right| \right)}}{\sin{\left(\frac{1}{2} \right)}}+C$$

Answer

$$$\int \frac{1}{- x \sin{\left(\frac{1}{2} \right)} + 1}\, dx = - \frac{\ln\left(\left|{x \sin{\left(\frac{1}{2} \right)} - 1}\right|\right)}{\sin{\left(\frac{1}{2} \right)}} + C$$$A

Integral of $$$\frac{1}{- x \sin{\left(\frac{1}{2} \right)} + 1}$$$

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