Evaluate the integral: I = integral dx/(sqrt((1+x)³(1-x)))
Evaluate the integral: $$ I = \int \frac{dx}{\sqrt{(1+x)^3(1-x)}}. $$
Explanation of Method
The half-angle-style substitution $x=\cos2\theta$ turns $1+x=2\cos^2\theta$ and $1-x=2\sin^2\theta$, collapsing the whole integrand to $-\sec^2\theta$, a direct antiderivative.
- Substitute $x=\cos2\theta$.
- Simplify using half-angle identities.
- Integrate $-\sec^2\theta$.
- Convert back to $x$.
Step 1 · Substitute
Let $x=\cos2\theta$, $0<\theta<\tfrac\pi2$, so $dx=-2\sin2\theta\,d\theta=-4\sin\theta\cos\theta\,d\theta$, and $1+x=2\cos^2\theta$, $1-x=2\sin^2\theta$:
$$ (1+x)^{3/2}\sqrt{1-x} = 2\sqrt2\cos^3\theta\cdot\sqrt2\sin\theta = 4\cos^3\theta\sin\theta. $$
Step 2 · Substitute into the integral
$$ I = \int\frac{-4\sin\theta\cos\theta\,d\theta}{4\cos^3\theta\sin\theta} = -\int\sec^2\theta\,d\theta. $$
Step 3 · Integrate
$$ I = -\tan\theta + C. $$
Step 4 · Convert back to $x$
From $\cos2\theta=x$: $\cos^2\theta=\dfrac{1+x}2$, $\sin^2\theta=\dfrac{1-x}2$, so:
$$ \tan\theta = \sqrt{\frac{1-x}{1+x}}. $$
$$ I = -\sqrt{\frac{1-x}{1+x}} + C = -\frac{\sqrt{1-x^2}}{1+x} + C. $$
Summary Table
| Quantity | Value |
|---|---|
| $\cos^2\theta$ | $(1+x)/2$ |
| $\sin^2\theta$ | $(1-x)/2$ |
Additional Notes
- The domain is $-1<x<1$.
- The result was verified symbolically: differentiating $-\sqrt{(1-x)/(1+x)}$ gives exactly $1/\sqrt{(1+x)^3(1-x)}$, confirmed numerically at $x=0$ and $x=1/2$.
Result
The solution to the integral is: