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Chapter 7: Problem 35

Evaluate the following integrals. $$\int \frac{d x}{\sqrt{3-2 x-x^{2}}}$$

Short Answer

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Question: Evaluate the integral $$\int \frac{d x}{\sqrt{3-2 x-x^{2}}}$$. Answer: $$\frac{1}{\sqrt{3}}\ln\left|\cfrac{\sqrt{3}}{\sqrt{3-2x-x^2}} + \cfrac{x-1}{\sqrt{3-2x-x^2}}\right| + C$$

Step by step solution

01

Identify R and h

The given integral is: $$\int \frac{d x}{\sqrt{3-2 x-x^{2}}}$$ Comparing it to the general expression \(\sqrt{R^2 - (x - h)^2}\), we can identify the radius R and the horizontal shift h. In our case, we have \(R^2 = 3\) and \((x - h)^2 = 2x + x^2\). From \(R^2 = 3\), we get \(R = \sqrt{3}\). For the second part, we can expand \((x - h)^2\): $$2x + x^2 = x^2 - 2hx + h^2$$ Equating the corresponding coefficients, we get: $$h^2 - 2hx + x^2 = 2x + x^2$$ $$h^2 - 2hx = 2x$$ From this equation, we find that \(h = 1\).
02

Apply trigonometric substitution

Substitute \(x = R\sin{\theta} + h\): $$x = \sqrt{3}\sin{\theta} + 1$$ Now, we differentiate with respect to \(\theta\): $$\frac{dx}{d\theta} = \sqrt{3}\cos{\theta}$$ Now replace \(x\) and \(dx\) in the integral: $$\int \frac{\sqrt{3}\cos{\theta}d\theta}{\sqrt{3 - 2(\sqrt{3}\sin{\theta} + 1) - (\sqrt{3}\sin{\theta} + 1)^2}}$$
03

Simplify the integrand

Simplify the expression inside the square root: $$\int \frac{\sqrt{3}\cos{\theta}d\theta}{\sqrt{3 - 2\sqrt{3}\sin{\theta} - 2 - 2\sqrt{3}\sin{\theta} - 3\sin^2{\theta}}}$$ $$\int \frac{\sqrt{3}\cos{\theta}d\theta}{\sqrt{1 - 4\sqrt{3}\sin{\theta} + 3\sin^2{\theta}}}$$ The expression inside the square root is now in the desired form. We can use the identity \(a^2\sin^2{\theta} + b^2\cos^2{\theta} = a^2 - b^2(a^2 - 1) = 1\): $$\int \frac{\sqrt{3}\cos{\theta}d\theta}{\sqrt{1^2 - (2\sqrt{3}\cos{\theta})^2}}$$
04

Evaluate the resulting integral

The simplified integral becomes: $$\int \frac{\sqrt{3}\cos{\theta}d\theta}{\sqrt{1 - 12\cos^2{\theta}}}$$ The integrand is now a standard result: $$\int \frac{\sqrt{3}\cos{\theta}d\theta}{\sqrt{1 - 12\cos^2{\theta}}} = \frac{1}{\sqrt{3}}\int \frac{d\theta}{\sin\theta}$$ $$= \frac{1}{\sqrt{3}}\ln|\csc{\theta} + \cot{\theta}| + C$$
05

Back-substitute x

Finally, we express the antiderivative in terms of x: $$\frac{1}{\sqrt{3}}\ln\left|\cfrac{1}{\sin(\arcsin(\frac{x-1}{\sqrt{3}}))} + \cfrac{\cos(\arcsin(\frac{x-1}{\sqrt{3}}))}{\sin(\arcsin(\frac{x-1}{\sqrt{3}}))}\right| + C$$ This can be further simplified to: $$\frac{1}{\sqrt{3}}\ln\left|\cfrac{\sqrt{3}}{\sqrt{3-2x-x^2}} + \cfrac{x-1}{\sqrt{3-2x-x^2}}\right| + C$$

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Most popular questions from this chapter

Use the following three identities to evaluate the given integrals. $$\begin{aligned}&\sin m x \sin n x=\frac{1}{2}[\cos ((m-n) x)-\cos ((m+n) x)]\\\&\sin m x \cos n x=\frac{1}{2}[\sin ((m-n) x)+\sin ((m+n) x)]\\\&\cos m x \cos n x=\frac{1}{2}[\cos ((m-n) x)+\cos ((m+n) x)]\end{aligned}$$ $$\int \sin 3 x \cos 7 x d x$$

The following integrals require a preliminary step such as long division or a change of variables before using partial fractions. Evaluate these integrals. $$\int \frac{e^{x}}{\left(e^{x}-1\right)\left(e^{x}+2\right)} d x$$

By reduction formula 4 in Section 3 $$\int \sec ^{3} u d u=\frac{1}{2}(\sec u \tan u+\ln |\sec u+\tan u|)+C$$ Graph the following functions and find the area under the curve on the given interval. $$f(x)=\left(9-x^{2}\right)^{-2},\left[0, \frac{3}{2}\right]$$

Use the indicated substitution to convert the given integral to an integral of a rational function. Evaluate the resulting integral. $$\int \frac{d x}{x \sqrt{1+2 x}} ; 1+2 x=u^{2}$$

An integrand with trigonometric functions in the numerator and denominator can often be converted to a rational integrand using the substitution \(u=\tan (x / 2)\) or \(x=2 \tan ^{-1} u .\) The following relations are used in making this change of variables. $$A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}$$ $$\text { Evaluate } \int \frac{d x}{1+\sin x+\cos x}$$
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