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Chapter 6: Problem 33

Evaluate the integral using (a) the given integration limits and (b) the limits obtained by trigonometric substitution. $$ \int_{0}^{\sqrt{3} / 2} \frac{t^{2}}{\left(1-t^{2}\right)^{3 / 2}} d t $$

Short Answer

Expert verified
The evaluated integral is \( \frac{1}{2} \)

Step by step solution

01

Method A: Direct Integration

Here, apply the given limits directly. A good substitution would be \(u = 1 - t^2\) which yields \(du = -2t dt\). When the lower limit \(t = 0\), we get \(u = 1\), and when the upper limit \(t = \sqrt{3}/2\), we get \(u = 1/4\). So, the integral changes to \(-\frac{1}{2} \int_1^{1/4} u^{-3/2} du\) which can be solved directly.
02

Calculus for Method A

This integral can be expressed as \(-\frac{1}{2} [-2u^{-1/2}]_1^{1/4} = - [(\frac{1}{4})^{1/2} - 1] = \(1 - \frac{1}{2}\) = \(\frac{1}{2}\). So, the direct method gives the result: \(\frac{1}{2}\)
03

Method B: Trigonometric Substitution

By using the trigonometric substitution, set \(t = \frac{\sin(\theta)}{2}\). Then \(dt = \frac{\cos(\theta)}{2} d\theta\). Now, we substitute these into the integral, yielding \(-\int_{0}^{\pi/3} \frac{\sin^{2}(\theta)}{4(cos^2(\theta))^{3/2}} cos(\theta) d\theta\), after substituting the limits. Now, using the property of cosines, we can simplify the integral.
04

Calculus for Method B

The integral now becomes \(-\int_{0}^{\pi/3} \frac{\sin^{2}(\theta)}{4} d\theta\). This integral can be solved using the power-reduction identity to give the result \(-[\frac{\theta}{4} - \frac{\sin(2\theta)}{8}]_{0}^{\pi/3}\), which also equals \(\frac{1}{2}\).
05

Comparison of the Results

Both methods gave the same result \(\frac{1}{2}\). This verifies that the methods are equivalent.

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