Evaluate the integral: I = integral dx/(sin x)
Evaluate the integral: $$ I = \int \frac{dx}{\sin x}. $$
Explanation of Method
This is the classical cosecant integral. Multiplying the integrand by $\dfrac{\csc x-\cot x}{\csc x - \cot x}$ turns the numerator into the derivative of the denominator, producing a standard logarithmic antiderivative.
- Rewrite $1/\sin x$ as $\csc x$.
- Multiply by the conjugate-style factor $(\csc x-\cot x)/(\csc x-\cot x)$.
- Recognize the numerator as the derivative of the denominator.
- Integrate using the standard logarithmic result.
Step 1 · Rewrite the integrand
The integrand $\dfrac1{\sin x}$ is the cosecant function:
$$ I = \int \csc x\,dx. $$
Step 2 · Apply the standard technique
Multiplying and dividing by $\csc x-\cot x$:
$$ I = \int \csc x\cdot\frac{\csc x-\cot x}{\csc x-\cot x}\,dx = \int \frac{\csc^2x-\csc x\cot x}{\csc x-\cot x}\,dx. $$
The numerator is exactly the derivative of the denominator, since $\dfrac{d}{dx}(\csc x-\cot x)=-\csc x\cot x+\csc^2x$.
Step 3 · Integrate
Letting $w=\csc x-\cot x$, so $dw=(\csc^2x-\csc x\cot x)\,dx$:
$$ I = \int \frac{dw}{w} = \ln|w| + C. $$
Step 4 · Substitute back
Substituting back $w=\csc x-\cot x$:
$$ I = \ln|\csc x-\cot x| + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Trick | multiply by $(\csc x-\cot x)/(\csc x-\cot x)$ |
| Substitution | $w = \csc x - \cot x$ |
| Antiderivative | $\ln|w|$ |
Additional Notes
- The domain excludes $x=k\pi$ for integer $k$, where $\sin x=0$.
- This is a memorized standard result — like $\int\sec x\,dx$ in Problem 113, the "multiply by a clever form of 1" trick is not something one would typically re-derive from scratch, but it is worth understanding once.
- The result can be checked by differentiating $\ln|\csc x-\cot x|$ with the chain rule and confirming it returns $\csc x = 1/\sin x$.
Result
The solution to the integral is: