Evaluate the integral: I = integral dx/(sqrt(1-x²)

Evaluate the integral: $$ I = \int \frac{dx}{\sqrt{1-x^2}\,\arcsin^5x}. $$

Explanation of Method

Since $1/\sqrt{1-x^2}$ is exactly the derivative of $\arcsin x$, substituting $u=\arcsin x$ reduces the integral to a plain power-rule form $\int u^{-5}\,du$.

Substitute $u=\arcsin x$.
Rewrite the integral in terms of $u$.
Apply the power rule for integration.
Substitute back and add the constant of integration.

Step 1 · Set up the substitution

Let:

$$ u = \arcsin x. $$

Then:

$$ du = \frac{dx}{\sqrt{1-x^2}}. $$

Step 2 · Rewrite the integral

Substituting into the integral:

$$ I = \int \frac{du}{u^5} = \int u^{-5}\,du. $$

Step 3 · Apply the power rule

Using $\int u^n\,du = \dfrac{u^{n+1}}{n+1}$ with $n=-5$:

$$ I = \frac{u^{-4}}{-4} + C = -\frac{1}{4u^4} + C. $$

Step 4 · Substitute back

Substituting back $u=\arcsin x$:

$$ I = -\frac{1}{4\arcsin^4x} + C. $$

Summary Table

Quantity	Value
Substitution	$u = \arcsin x$
$dx/\sqrt{1-x^2}$ in terms of $du$	$du$
Antiderivative	$-1/(4u^4)$

Additional Notes

The domain requires $x\in(-1,1)$ with $x\neq0$, since $\arcsin0=0$ makes the integrand undefined there.
This is the power-rule counterpart to Problem 106's logarithmic case: whenever $\arcsin x$ appears raised to a power $n\neq-1$ alongside its own derivative, the power rule applies instead of the logarithm.
The result can be checked by differentiating $-\tfrac{1}{4}(\arcsin x)^{-4}$ with the chain rule and confirming it returns $\dfrac{1}{\sqrt{1-x^2}\arcsin^5x}$.

Result

The solution to the integral is:

$$ I = -\frac{1}{4\arcsin^4x} + C. $$