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Chapter 1: Problem 2

Evaluate the following integrals: $$ \int \frac{d x}{\left(4 x-x^{2}\right)^{3 / 2}} $$

Short Answer

Expert verified
Question: Evaluate the integral $\int \frac{dx}{(4x - x^2)^{3/2}}$. Answer: $\frac{1}{4} \left(\arcsin{\frac{x - 2}{2}} - \frac{\sqrt{4 - (x - 2)^2}}{2}\right) + C$

Step by step solution

01

Complete the square in the denominator

To complete the square, rewrite the quadratic function in the denominator as the square of a binomial. We can rewrite the expression \(4x - x^2\) in the following manner: $$ -(x^2 - 4x) = -(x^2 - 4x + 4 - 4) = -(x - 2)^2 + 4 $$ Substitute this back into the original integral to obtain: $$ \int \frac{dx}{\left(-(x - 2)^2 + 4\right)^{3/2}} $$
02

Apply the trigonometric substitution

A suitable trigonometric substitution for this problem is \(x - 2 = 2\sin{t}\). Differentiating both sides with respect to t, we get: $$ dx = 2\cos{t}\,dt $$ The integral now becomes: $$ \int \frac{2\cos{t}\,dt}{\left(-4\sin^2{t} + 4\right)^{3/2}} $$ Simplify the integrand: $$ \int \frac{2\cos{t}\, dt}{(4\cos^2{t})^{3/2}} = \int \frac{2\cos{t}\, dt}{8\cos^3{t}} $$ Canceling out the common factors, the integral simplifies to: $$ \int \frac{dt}{4\cos^2{t}} $$
03

Evaluate the integral

The remaining integral can be solved using the Pythagorean identity \(\sin^2{t} + \cos^2{t} = 1\), so we get: $$ \int \frac{dt}{4(1 - \sin^2{t})} $$ Using the double angle identity \(\frac{1}{1-\sin{t}} = \frac{1+\sin{t}}{1-\sin^{2}{t}} \), $$ \int \frac{1+\sin{t}}{4} dt $$ Now we can integrate: $$ \frac{1}{4}\left(\int dt + \int \sin t\, dt\right) = \frac{1}{4}\left(t-\cos{t}\right) + C $$
04

Replace the trigonometric function with the original variable

Lastly, we need to write iout answer back in terms of the original variable x. Recall that we used the substitution \(x - 2 = 2\sin{t}\). From this, obtain: $$ \sin{t} = \frac{x - 2}{2} $$ Now, using the Pythagorean theorem, we have: $$ \cos{t} = \sqrt{1 - \sin^2{t}} = \sqrt{1 - \left(\frac{x - 2}{2}\right)^2} = \sqrt{1 - \frac{(x - 2)^2}{4}} = \frac{\sqrt{4 - (x - 2)^2}}{2} $$ Substitute this result back into the integral we obtained earlier: $$ \frac{1}{4} \left(t-\frac{\sqrt{4 - (x - 2)^2}}{2}\right) + C = \frac{1}{4} \left(\arcsin{\frac{x - 2}{2}} - \frac{\sqrt{4 - (x - 2)^2}}{2}\right) + C $$ This is the final answer for the given integral.

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Most popular questions from this chapter

Evaluate the following integrals: (i) \(\int \frac{5 x^{2}-12}{\left(x^{2}-6 x+13\right)^{2}} d x\) (ii) \(\int \frac{x^{3}+x-1}{\left(x^{2}+2\right)^{2}} d x\) (iii) \(\int \frac{x^{6}+x^{4}-4 x^{2}-2}{x^{3}\left(x^{2}+1\right)^{2}} d x\) (iv) \(\int \frac{d x}{x^{4}\left(x^{3}+1\right)^{2}}\)

(i) There are two values of a for which \(\int \sqrt{1+a \sin ^{2} \theta} d \theta\) is elementary. What are they? (ii) From (1) deduce that there are two values of a for which \(\int \frac{\sqrt{1+a x^{2}}}{\sqrt{1-x^{2}}} \mathrm{dx}\) is elementary.

Evaluate the following integrals: (i) \(\int \frac{\mathrm{dx}}{\left(\mathrm{x}^{2}+1\right) \sqrt{\mathrm{x}}}\) (ii) \(\int \frac{\mathrm{dx}}{\left(\mathrm{x}^{2}+5 \mathrm{x}+6\right) \sqrt{\mathrm{x}+1}}\) (iii) \(\int \frac{d x}{\left(x^{2}-4\right) \sqrt{x+1}}\)

Evaluate the following integrals : $$ \int x^{-1}\left(1+x^{1 / 3}\right)^{-3} d x $$

Evaluate the following integrals: (i) \(\int \frac{2 x+\sin 2 x}{1+\cos 2 x} d x\) (ii) \(\int\left(\tan (\ln x)+\sec ^{2}(\ln x)\right\\} d x\) (iii) \(\int \frac{x+\sqrt{\left(1-x^{2}\right)} \sin ^{-1} x}{\sqrt{\left(1-x^{2}\right)}} d x\)
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