/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 32 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 32

Evaluate the following integrals. $$\int \sqrt{9-4 x^{2}} d x$$

Short Answer

Expert verified
Question: Evaluate the following integral: \(\int \sqrt{9-4 x^{2}} d x\). Answer: The evaluated integral is \(\int \sqrt{9-4 x^{2}} d x = 9x - \frac{9}{4}x^3 + C\).

Step by step solution

01

Make a substitution

Let's make the following substitution: $$x = \frac{3}{2} \sin \theta$$ The differential dx can be found by differentiating with respect to \(\theta\): $$d x = \frac{3}{2} \cos \theta\ d\theta$$ Now, substitute x into the integrand and replace dx with their expressions in terms of d胃.
02

Replace x and dx

Replace x and dx in the integral: $$\int \sqrt{9 - 4x^2} d x = \int \sqrt{ 9 - 4 \left(\frac{3}{2} \sin \theta\right)^{2}} \left(\frac{3}{2}\cos \theta\ d\theta\right)$$
03

Simplify the integrand

Now, simplify the integrand: \begin{align*} \int \sqrt{ 9 - 4 \left(\frac{3}{2} \sin \theta\right)^{2}} \left(\frac{3}{2}\cos \theta\ d\theta\right) &= \int \sqrt{ 9 - 9\sin^2\theta} \left(\frac{3}{2}\cos \theta\ d\theta\right)\\ &= \int \sqrt{ 9(1 - \sin^2\theta)}\frac{3}{2}\cos\theta d\theta \\ &= \int \sqrt{ 9\cos^{2}\theta} \cdot \frac{3}{2}\cos\theta d\theta \\ &= \int 3\cos^{2}\theta\cdot \frac{3}{2}\cos\theta d\theta \\ &= \frac{9}{2} \int \cos^3\theta\ d\theta \\ \end{align*}
04

Evaluate the integral

Now, we will evaluate the integral: $$\frac{9}{2} \int \cos^3\theta\ d\theta$$ To evaluate the above integral, use the following expression: $$\cos^3\theta= (1 - \sin^2\theta)\cos\theta$$ Replace cos鲁胃 with the expression above and integrate: $$\frac{9}{2} \int (1 - \sin^2\theta)\cos\theta\ d\theta$$ Now, perform integration by substitution: let \(u = \sin\theta\), thus \(du = \cos\theta\ d\theta\): $$\frac{9}{2} \int (1 - u^2)du = \frac{9}{2}\left(u - \frac{u^3}{3}\right) + C$$ Now, replace u with sin胃: $$\frac{9}{2}\left(\sin\theta - \frac{\sin^3\theta}{3}\right) + C = \frac{9}{2}\left(\frac{3\sin\theta - \sin^{3}\theta}{3}\right) + C$$
05

Reverse the substitution

Now, reverse the substitution, replacing \(\sin\theta\) with the expression in terms of x: $$\frac{9}{2}\left(\frac{3\frac{3}{2} \sin\theta - (\frac{3}{2}\sin\theta)^{3}}{3}\right) + C$$ Recall that \(x = \frac{3}{2}\sin\theta\), thus \(\sin\theta = \frac{2}{3}x\): $$\frac{9}{2}\left( \frac{6x - (\frac{3}{2} \cdot x^3)}{3} \right) + C = \frac{9}{2}\left(2x-\frac{1}{2}x^{3}\right) + C$$ Simplify the expression: $$9x - \frac{9}{4}x^3 + C$$ The evaluated integral is: $$\int \sqrt{9-4 x^{2}} d x = 9x - \frac{9}{4}x^3 + C$$

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Most popular questions from this chapter

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