Evaluate the integral: I = integral (x cos x)/(sin³x) dx
Evaluate the integral: $$ I = \int \frac{x\cos x}{\sin^3x}\,dx. $$
Explanation of Method
Since $\cos x/\sin^3x$ has an antiderivative in $x$ alone (via $u=\sin x$), integration by parts with $u=x$ and $dv=\cos x/\sin^3x\,dx$ leaves a leftover $\csc^2x$ integral, which is standard.
- Find $v=\int \cos x/\sin^3x\,dx$ via the substitution $w=\sin x$.
- Apply integration by parts with $u=x$, $dv=\cos x/\sin^3x\,dx$.
- Integrate the leftover $\csc^2x$ term.
Step 1 · Find $v$
Let $w=\sin x$, $dw=\cos x\,dx$. Then:
$$ v = \int \frac{\cos x}{\sin^3x}\,dx = \int w^{-3}\,dw = -\frac{1}{2w^2} = -\frac{1}{2\sin^2x}. $$
Step 2 · Set up integration by parts
Let:
$$ u = x \quad \Rightarrow \quad du = dx, \qquad dv = \frac{\cos x}{\sin^3x}\,dx \quad \Rightarrow \quad v = -\frac{1}{2\sin^2x}. $$
Applying the formula:
$$ I = -\frac{x}{2\sin^2x} + \frac12\int \frac{dx}{\sin^2x}. $$
Step 3 · Integrate the leftover term
Using $\int \csc^2x\,dx = -\cot x$:
$$ I = -\frac{x}{2\sin^2x} + \frac12\left(-\cot x\right) + C. $$
Step 4 · Simplify
$$ I = -\frac{x}{2\sin^2x} - \frac12\cot x + C. $$
Summary Table
| Quantity | Value |
|---|---|
| $u$ | $x$ |
| $v=\int\cos x/\sin^3x\,dx$ | $-1/(2\sin^2x)$ |
| Leftover integral | $\tfrac12\int\csc^2x\,dx=-\tfrac12\cot x$ |
Additional Notes
- The domain excludes $x=k\pi$ for integer $k$, where $\sin x=0$.
- Finding $v$ first via a plain substitution (before ever invoking integration by parts) is a useful strategy whenever the "non-$x$" part of the integrand, like $\cos x/\sin^3x$ here, is easy to integrate on its own.
- The result can be checked by differentiating $-\tfrac{x}{2\sin^2x}-\tfrac12\cot x$ and confirming it returns $x\cos x/\sin^3x$.
Result
The solution to the integral is: