Evaluate the integral: I = integral (cos³ x)/(sin x) dx
Evaluate the integral: $$ I = \int \frac{\cos^3 x}{\sin x} \, dx. $$
Explanation of Method
Splitting off one factor of $\cos x$ and rewriting the remaining $\cos^2x$ via the Pythagorean identity separates the integrand into a $\cot x$ term and a $\sin x\cos x$ term, each of which is a standard or simple substitution integral.
- Split off one factor of $\cos x$ and rewrite $\cos^2x$ using the Pythagorean identity.
- Separate the integrand into two simpler terms.
- Integrate each term using substitution.
- Combine the results and add the constant of integration.
Step 1 · Rewrite the integrand
The expression involves powers of $\cos x$ and $\sin x$, and it suggests that we can use trigonometric identities to simplify the integral. To begin, we can rewrite $\cos^3x$ as:
$$ \cos^3x = \cos x\cdot \cos^2x. $$
Now, we know that $\cos^2x = 1-\sin^2x$, so we substitute:
$$ I = \int \frac{\cos x\,(1-\sin^2x)}{\sin x}\,dx. $$
Step 2 · Separate into two terms
This separates into two terms:
$$ I = \int \left(\frac{\cos x}{\sin x} - \frac{\cos x\sin^2x}{\sin x}\right) dx. $$
Simplifying each term:
$$ I = \int \frac{\cos x}{\sin x}\,dx - \int \cos x\sin x\,dx. $$
Step 3 · Evaluate each integral
First integral: $\displaystyle \int \frac{\cos x}{\sin x}\,dx$. We use the substitution $u=\sin x$, so that $du=\cos x\,dx$. The integral becomes:
$$ \int \frac{du}{u} = \ln|u| + C = \ln|\sin x| + C_1. $$
Second integral: $\displaystyle \int \cos x\sin x\,dx$. This is a straightforward trigonometric integral. We use the substitution $u=\sin x$, so $du=\cos x\,dx$. The integral becomes:
$$ \int u\,du = \frac{u^2}{2} + C = \frac{\sin^2x}{2} + C_2. $$
Step 4 · Combine the results
Now, combining both integrals:
$$ I = \ln|\sin x| - \frac{\sin^2x}{2} + C, $$
where $C$ is the constant of integration.
Summary Table
| Term | Antiderivative |
|---|---|
| $\cos x/\sin x$ | $\ln|\sin x|$ |
| $-\sin x\cos x$ | $-\sin^2x/2$ |
Additional Notes
- The domain of $I$ excludes $x=k\pi$ for integer $k$, where $\sin x=0$.
- Both resulting integrals happen to use the same substitution $u=\sin x$, but produce different antiderivative shapes ($\ln|u|$ versus $u^2/2$) because the first is a logarithmic pattern and the second is a straightforward power-rule pattern.
- The result can be checked by differentiating $\ln|\sin x| - \tfrac{\sin^2x}{2}$ and confirming it returns $\dfrac{\cos^3x}{\sin x}$.
Result
The solution to the integral is: