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Chapter 5: Q. 55 (page 465)

Solve the integral $鈭� \frac{3 + x}{\sqrt{9 - 4 x^{2}}} d x$

Short Answer

Expert verified

$- \frac{1}{4} 鈭� \sqrt{(- 4 x^{2} + 9)} + \frac{3}{2} \sin^{- 1} (\frac{2 x}{3}) + c$

Step by step solution

01

Step 1. Given

$鈭� \frac{3 + x}{\sqrt{9 - 4 x^{2}}} d x$

02

Step 3. Calculation

$鈭� \frac{3 + x}{\sqrt{9 - 4 x^{2}}} d x = 鈭� \frac{x}{\sqrt{9 - 4 x^{2}}} d x + 鈭� \frac{3}{\sqrt{9 - 4 x^{2}}} d x = - \frac{1}{8} 鈭� \frac{1}{\sqrt{u}} d x + 鈭� \frac{3}{\sqrt{9 - 4 x^{2}}} d x [u = - 4 x^{2} + 9, d u = - 8 x d x] = - \frac{2 \sqrt{u}}{8} + 鈭� \frac{3}{\sqrt{9 - 4 x^{2}}} d x = - \frac{\sqrt{u}}{4} + 鈭� \frac{3}{\sqrt{9 - 4 x^{2}}} d x = - \frac{1}{4} \sqrt{(- 4 x^{2} + 9} + 鈭� \frac{3}{\sqrt{9 - 4 x^{2}}} d x = - \frac{1}{4} \sqrt{(- 4 x^{2} + 9} + 3 鈭� \frac{1}{2} d u = - \frac{1}{4} \sqrt{(- 4 x^{2} + 9} + \frac{3}{2} (u) = - \frac{1}{4} \sqrt{(- 4 x^{2} + 9} + \frac{3}{2} (\sin^{- 1} (\frac{2 x}{3})) + C$

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Most popular questions from this chapter

Solve $鈭� \frac{1}{x^{2} + 9} d x$ the following two ways:

(a) with the trigonometric substitution x = 3 tan u;

(b) with algebra and the derivative of the arctangent.

Solve given definite integrals.

${鈭�}_{0}^{4} x \sqrt{x^{2} + 4} d x$

Complete the square for each quadratic in Exercises 28鈥�33. Then describe the trigonometric substitution that would be appropriate if you were solving an integral that involved that quadratic.

$x^{2} - 5 x + 1$

Solve the integral: $鈭� 3 x e^{x^{2}} d x$

Give an example of an integral for which trigonometric substitution is possible but an easier method is available. Then give an example of an integral that we still don鈥檛 know how to solve given the techniques we know at this point.

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