Evaluate the integral: I = integral x³sqrt((1+x²)/(1-x²))
Evaluate the integral: $$ I = \int x^3\sqrt{\frac{1+x^2}{1-x^2}}\,dx. $$
Explanation of Method
The substitution $x=\sin\theta$ turns $1-x^2$ into $\cos^2\theta$, reducing the radical to $\sqrt{1+\sin^2\theta}/\cos\theta$; the $\cos\theta$ cancels against $dx$, leaving $\int\sin^3\theta\sqrt{1+\sin^2\theta}\,d\theta$, which a further substitution $u=\cos\theta$ reduces to polynomial-times-radical integrals.
- Substitute $x=\sin\theta$.
- Substitute again $u=\cos\theta$.
- Integrate using standard radical formulas.
- Convert back to $x$.
Step 1 · First substitution
Let $x=\sin\theta$, $-\tfrac\pi2<\theta<\tfrac\pi2$, so $dx=\cos\theta\,d\theta$ and $\sqrt{(1+x^2)/(1-x^2)}=\sqrt{1+\sin^2\theta}/\cos\theta$:
$$ I = \int\sin^3\theta\sqrt{1+\sin^2\theta}\,d\theta. $$
Step 2 · Second substitution
Let $u=\cos\theta$, $\sin^2\theta=1-u^2$, $\sin\theta\,d\theta=-du$, and $1+\sin^2\theta=2-u^2$:
$$ I = -\int\left(1-u^2\right)\sqrt{2-u^2}\,du = -\int\sqrt{2-u^2}\,du + \int u^2\sqrt{2-u^2}\,du. $$
Step 3 · Integrate
Using $\int\sqrt{2-u^2}\,du=\tfrac u2\sqrt{2-u^2}+\arcsin\left(\tfrac u{\sqrt2}\right)$ and $\int u^2\sqrt{2-u^2}\,du=-\tfrac{u(2-u^2)^{3/2}}4+\tfrac12\int\sqrt{2-u^2}\,du$:
$$ I = -\frac{u\sqrt{2-u^2}(3-u^2)}4 - \frac12\arcsin\left(\frac u{\sqrt2}\right) + C. $$
Step 4 · Convert back to $x$
Since $u=\cos\theta=\sqrt{1-x^2}$ and $2-u^2=1+x^2$:
$$ I = -\frac{(2+x^2)\sqrt{1-x^4}}4 - \frac12\arcsin\sqrt{\frac{1-x^2}2} + C. $$
Summary Table
| Piece | Antiderivative |
|---|---|
| $-\sqrt{2-u^2}$ | $-\tfrac u2\sqrt{2-u^2}-\arcsin(u/\sqrt2)$ |
| $u^2\sqrt{2-u^2}$ | $-\tfrac{u(2-u^2)^{3/2}}4+\tfrac12\left[\tfrac u2\sqrt{2-u^2}+\arcsin(u/\sqrt2)\right]$ |
Additional Notes
- The domain requires $-1<x<1$.
- The result was verified by differentiation: writing $g=2+x^2$, $h=\sqrt{1-x^4}$, and $p=\sqrt{(1-x^2)/2}$, combining $\tfrac{d}{dx}[-gh/4]$ and $\tfrac{d}{dx}[-\tfrac12\arcsin p]$ over the common denominator $\sqrt{1-x^4}$ gives numerator $x^3(x^2+1)$, and $x^3(x^2+1)/\sqrt{1-x^4}=x^3\sqrt{(1+x^2)/(1-x^2)}$, matching the integrand exactly.
Result
The solution to the integral is: