# What is f(x) = int 3x-2cscx dx if f((5pi)/4) = 0 ?

Jan 11, 2018

$f \left(x\right) = \frac{3}{2} {x}^{2} + 2 \ln | \csc \left(x\right) + \cot \left(x\right) | - 21.3691$

#### Explanation:

$\int 3 x - 2 \csc x \mathrm{dx}$

$= \int 3 x \mathrm{dx} - 2 \int \csc x \mathrm{dx}$

To integrate the second integral re write and use the following substitution:

$\int \csc x \mathrm{dx} = \int \csc x \frac{\csc x + \cot x}{\csc x + \cot x} \mathrm{dx}$

$u = \csc x + \cot x$
$\to \mathrm{du} = - \cot x \csc x - {\csc}^{2} x \mathrm{dx} = - \csc x \left(\csc x + \cot x\right) \mathrm{dx}$

Now substitute into the integral:

$\int \csc x \frac{\csc x + \cot x}{\csc x + \cot x} \mathrm{dx} = - \int \frac{\mathrm{du}}{u} = - \ln u + {C}_{0}$

Reverse the substitution:

$- \ln | u | + {C}_{0} = - \ln | \csc \left(x\right) + \cot \left(x\right) | + {C}_{0}$

So, returning to the original integral:

$\int 3 x \mathrm{dx} - 2 \int \csc x \mathrm{dx} = \frac{3}{2} {x}^{2} + 2 \ln | \csc \left(x\right) + \cot \left(x\right) | + C$

Now solve for $C$.

$f \left(\frac{5 \pi}{4}\right)$
$= \frac{3}{2} {\left(\frac{5 \pi}{4}\right)}^{2} + 2 \ln | \csc \left(\frac{5 \pi}{4}\right) + \cot \left(\frac{5 \pi}{4}\right) | + C = 0$

$\frac{75}{32} {\pi}^{2} + 2 \ln \left(- \sqrt{2} + 1\right) + C$

$\to 21.3691 + C = 0 \to C = - 21.3691$

$f \left(x\right) = \frac{3}{2} {x}^{2} + 2 \ln | \csc \left(x\right) + \cot \left(x\right) | - 21.3691$