Evaluate the integral: I = integral (sin x dx)/(sqrt(cos³
Evaluate the integral: $$ I = \int \frac{\sin x\,dx}{\sqrt{\cos^3 x}}. $$
Explanation of Method
The derivative of $\cos x$ is $-\sin x$, which matches the numerator up to sign. Substituting $u=\cos x$ reduces the integral to a plain power-rule form $\int u^{-3/2}\,du$.
- Rewrite the radical as a fractional power.
- Substitute $u=\cos x$.
- Apply the power rule for integration.
- Substitute back and add the constant of integration.
Step 1 · Rewrite the denominator
The square root of $\cos^3x$ can be written as a fractional power:
$$ \sqrt{\cos^3x} = \cos^{3/2}x. $$
Thus, the integral becomes:
$$ I = \int \frac{\sin x\,dx}{\cos^{3/2}x}. $$
Step 2 · Substitution
Let:
$$ u = \cos x \quad \text{so that} \quad du = -\sin x\,dx. $$
Substituting into the integral:
$$ I = -\int \frac{du}{u^{3/2}} = -\int u^{-3/2}\,du. $$
Step 3 · Apply the power rule
Using $\int u^n\,du = \dfrac{u^{n+1}}{n+1}$ with $n=-\tfrac32$:
$$ -\int u^{-3/2}\,du = -\frac{u^{-1/2}}{-1/2} + C = 2u^{-1/2} + C. $$
Step 4 · Substitute back
Substituting back $u=\cos x$:
$$ I = \frac{2}{\sqrt{\cos x}} + C = 2\sec^{1/2}x + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $u = \cos x$ |
| $\sin x\,dx$ in terms of $du$ | $-du$ |
| Antiderivative | $2u^{-1/2}$ |
Additional Notes
- The integrand requires $\cos x>0$, since it appears raised to a non-integer power inside the antiderivative.
- This is the mirror image of Problem 101: here the numerator matches (up to sign) the derivative of $\cos x$ instead of $\sin x$, leading to the same power-rule pattern.
- The result can be checked by differentiating $2\cos^{-1/2}x$ with the chain rule and confirming it returns $\sin x/\sqrt{\cos^3x}$.
Result
The solution to the integral is: