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Chapter 1: Problem 3

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

Short Answer

Expert verified
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

01

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: $$\int \frac{d x}{\sqrt{\left(2 x-x^{2}\right)^{3}}} = \int \frac{2u\ du}{(-2x+2)\cdot u^3}$$
02

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: $$ \int \frac{2u\ du}{(-2x+2)\cdot u^3} = \int \frac{2u\ du}{(-2(2 - u^2)+2)u^3} = \int \frac{2u\ du}{(2u^2)u^3}$$
03

Simplifying the resulting integral

Now we can simplify the integral as follows: $$ \int \frac{2u\ du}{(2u^2)u^3} = \int \frac{1}{u^4} du = \int u^{-4} du$$
04

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: $$2x - x^2 = u^2 \implies u^3 = \sqrt[3]{(2x-x^2)^3}$$
05

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

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Most popular questions from this chapter

Evaluate the following integrals : (i) \(\int \frac{(\sqrt{x}+1)\left(x^{2}-\sqrt{x}\right)}{x \sqrt{x}+x+\sqrt{x}} d x\) (ii) \(\int \frac{\sqrt{1-\mathrm{x}^{2}}-1}{\mathrm{x}}\left(\frac{1-\mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}+\mathrm{x}-1}+\frac{\sqrt{1+\mathrm{x}}}{\sqrt{1+\mathrm{x}}-\sqrt{1-\mathrm{x}}}\right) \mathrm{dx}\) (iii) \(\int \frac{x^{4}+5 x^{3}+15 x-9}{\frac{x^{6}+3 x^{4}}+\frac{9}{x^{4}}}{\left(x^{3}-4 x+3 x^{2}-12\right) / x^{5}} d x\) (iv) \(\int \frac{\sqrt[3]{x+\sqrt{2-x^{2}}} \sqrt[6]{1-x \sqrt{2-x^{2}}}}{\sqrt[3]{1-x^{2}}} d x\)

Evaluate the following integrals: (i) \(\int \mathrm{e}^{\mathrm{x}} \frac{\left(\mathrm{x}^{2}-3 \mathrm{x}+3\right)}{(\mathrm{x}+2)^{2}} \mathrm{dx}\) (ii) \(\int \frac{\mathrm{e}^{\mathrm{x}}\left(\mathrm{x}^{2}+1\right)}{(\mathrm{x}+1)^{2}} \mathrm{dx}\) (iii) \(\int \mathrm{e}^{x} \frac{(1-x)^{2}}{\left(1+x^{2}\right)^{2}} d x\) (iv) \(\int \frac{x^{2} e^{x}}{(x+2)^{2}} d x\)

Evaluate the following integrals : $$\int \frac{\mathrm{dx}}{\sqrt{1-\mathrm{x}^{2}}-1}$$

Evaluate the following integrals : $$ \int \sqrt[3]{1+\sqrt[4]{x}} d x $$

Only one of the functions \(\sqrt{x} \sqrt[3]{1-x}\) and \(\sqrt{1-x} \sqrt[3]{1-x}\) has an elementary antiderivative. Find the function.
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