Question

Evaluate the following integrals : $$\int \frac{d x}{\sqrt{\left(2 x-x^{2}\right)^{3}}}$$

Short answer

Answer: The evaluated integral is $$\int \frac{d x}{\sqrt{\left(2 x-x^{2}\right)^{3}}} = -\frac{1}{3\sqrt[3]{(2x-x^2)^3}} + C$$.

Step-by-step solution

Simplifying the integral using the substitution method

Let's substitute \(2x - x^2 = u^2\). Therefore, differentiating both sides with respect to \(x\), we get: $$-2x + 2 = 2u \frac{du}{dx}$$ Now, solve for \(dx\): $$ dx = \frac{2u \ du }{-2x + 2}$$ Replace \(dx\) and \((2x-x^2)\) by using the substitution in the integral: $$\int \frac{d x}{\sqrt{\left(2 x-x^{2}\right)^{3}}} = \int \frac{2u\ du}{(-2x+2)\cdot u^3}$$

Eliminating x using the substitution

To eliminate x in the integral, rewrite the substitution \(2x - x^2 = u^2\) as: $$x = 2 - u^2$$ Now substitute x into the integral: $$ \int \frac{2u\ du}{(-2x+2)\cdot u^3} = \int \frac{2u\ du}{(-2(2 - u^2)+2)u^3} = \int \frac{2u\ du}{(2u^2)u^3}$$

Simplifying the resulting integral

Now we can simplify the integral as follows: $$ \int \frac{2u\ du}{(2u^2)u^3} = \int \frac{1}{u^4} du = \int u^{-4} du$$

Evaluate the simplified integral

Now we can evaluate the integral: $$\int u^{-4} du = -\frac{1}{3u^3} + C$$ We still need to express the result in terms of \(x\). Remember the substitution we made earlier: $$2x - x^2 = u^2 \implies u^3 = \sqrt[3]{(2x-x^2)^3}$$

Write the result in terms of x

Replace \(u^3\) with the expression involving \(x\) to obtain the final result: $$-\frac{1}{3u^3} + C = -\frac{1}{3\sqrt[3]{(2x-x^2)^3}} + C$$ Thus, the evaluated integral is: $$\int \frac{d x}{\sqrt{\left(2 x-x^{2}\right)^{3}}} = -\frac{1}{3\sqrt[3]{(2x-x^2)^3}} + C$$