# What is the antiderivative of 1/sinx?

Sep 19, 2016

It is $- \ln \left\mid \csc x + \cot x \right\mid$

#### Explanation:

$\frac{1}{\sin} x = \csc x = \csc x \frac{\csc x + \cot x}{\csc x + \cot x}$

$= \frac{{\csc}^{2} x + \csc x \cot x}{\csc x + \cot x}$

The numerator is the opposite (the 'negative') of the derivative of the denomoinator.

So the antiderivative is minus the natural logarithm of the denominator.

$- \ln \left\mid \csc x + \cot x \right\mid$.

(If you've learned the technique of substitution, we can use $u = \csc x + \cot x$, so $\mathrm{du} = - {\csc}^{2} x - \csc x \cot x$. The expression becomes $- \frac{1}{u} \mathrm{du}$.)

You can verify this answer by differentiating.

May 7, 2018

A different approach to it

$\int \frac{1}{\sin} x \mathrm{dx}$ $=$

$\int \sin \frac{x}{\sin} ^ 2 x \mathrm{dx}$

$\int \sin \frac{x}{1 - {\cos}^{2} x} \mathrm{dx}$

Substitute

$\cos x = u$

$- \sin x \mathrm{dx} = \mathrm{du}$

$\sin x \mathrm{dx} = - \mathrm{du}$

$=$ $- \int \frac{1}{1 - {u}^{2}} \mathrm{du}$

• $\frac{1}{1 - {u}^{2}} = \frac{1}{\left(u - 1\right) \left(u + 1\right)} = \frac{A}{u - 1} + \frac{B}{u + 1}$ $=$

$\frac{A \left(u + 1\right) + B \left(u - 1\right)}{\left(u - 1\right) \left(u + 1\right)}$

We need $A \left(u + 1\right) + B \left(u - 1\right) = 1$ $\iff$

$A u + A + B u - B = 1$ $\iff$

$\left(A + B\right) u + A - B = 1$ $\iff$

$\left(A + B\right) u + A - B = 0 u + 1$ $\iff$

$\left\{\begin{matrix}A + B = 0 \text{ " \\ A-B=1" }\end{matrix}\right.$ $\iff$

$\left\{\begin{matrix}A + B = 0 \text{ " \\ A=B+1" }\end{matrix}\right.$ $\iff$

$\left\{\begin{matrix}B + 1 + B = 0 \text{ " \\ A=B+1" }\end{matrix}\right.$ $\iff$

$\left\{\begin{matrix}B = - \frac{1}{2} \text{ " \\ A=1/2" }\end{matrix}\right.$

Therefore, $- \int \frac{1}{1 - {u}^{2}} \mathrm{du}$ $=$

$- \int \left(\frac{\frac{1}{2}}{u - 1} - \frac{\frac{1}{2}}{u + 1}\right) \mathrm{du}$ $=$

$\frac{1}{2} \int \left(\frac{1}{u + 1} - \frac{1}{u - 1}\right) \mathrm{du}$ $=$

$\frac{1}{2} \int \left(\frac{\left(u + 1\right) '}{u + 1} - \frac{\left(u - 1\right) '}{u - 1}\right) \mathrm{du}$ $=$

$\frac{1}{2} \left(\ln | u + 1 | - \ln | u - 1 | + c\right)$ $=$

$\frac{1}{2} \left(\ln | \frac{u + 1}{u - 1} | + c\right)$ $=$

$\frac{1}{2} \left(\ln | \frac{\cos x + 1}{\cos x - 1} | + c\right)$ $=$

$\frac{1}{2} \left(\ln | \frac{1 - \cos x}{1 + \cos x} | + c\right)$

$\ln | \tan \left(\frac{x}{2}\right) | + c '$ ,

$\left(c , c '\right)$$\in$$\mathbb{R}$