Evaluate the integral: I = integral (cos x dx)/(cube root
Evaluate the integral: $$ I = \int \frac{\cos x\,dx}{\sqrt[3]{\sin^2 x}}. $$
Explanation of Method
The derivative of $\sin x$ is $\cos x$, which appears exactly in the numerator, while the denominator is a power of $\sin x$. This makes $u=\sin x$ the natural substitution, reducing the integral to a plain power-rule form.
- Rewrite the radical as a fractional power.
- Substitute $u=\sin x$.
- Apply the power rule for integration.
- Substitute back and add the constant of integration.
Step 1 · Rewrite the denominator
The cube root of $\sin^2x$ can be written as a fractional power:
$$ \sqrt[3]{\sin^2x} = (\sin^2x)^{1/3} = \sin^{2/3}x. $$
Thus, the integral becomes:
$$ I = \int \frac{\cos x\,dx}{\sin^{2/3}x}. $$
Step 2 · Substitution
Let:
$$ u = \sin x \quad \text{so that} \quad du = \cos x\,dx. $$
Substituting into the integral:
$$ I = \int \frac{du}{u^{2/3}} = \int u^{-2/3}\,du. $$
Step 3 · Apply the power rule
Using $\int u^n\,du = \dfrac{u^{n+1}}{n+1}$ with $n=-\tfrac23$:
$$ I = \frac{u^{1/3}}{1/3} + C = 3u^{1/3} + C. $$
Step 4 · Substitute back
Substituting back $u=\sin x$:
$$ I = 3\sin^{1/3}x + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $u = \sin x$ |
| $\cos x\,dx$ in terms of $du$ | $du$ |
| Antiderivative | $3u^{1/3}$ |
Additional Notes
- The integrand is undefined where $\sin x = 0$, i.e. at $x=k\pi$ for integer $k$.
- Any integral where the numerator is exactly the derivative of a power of a function in the denominator collapses to this same power-rule pattern — no trigonometric identities are needed.
- The result can be checked by differentiating $3\sin^{1/3}x$ with the chain rule and confirming it returns $\cos x/\sqrt[3]{\sin^2x}$.
Result
The solution to the integral is: