Q 19. Consider the integral 鈭玸in2x聽... [FREE SOLUTION]

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Chapter 5: Q 19. (page 451)

Consider the integral $鈭� \sin^{2} x \cos^{2} x d x$ . We solved this integral in Example 2(b) by applying double-angle identities at the very beginning.
(a) Solve this integral by applying the identity $\sin^{2} x = 1 - \cos^{2} x$
(b) Solve this integral another way, by applying the identity $\sin x \cos x = \frac{1}{2} \sin 2 x$

Short Answer

Expert verified

(a) $\begin{array}{rcl} \frac{x}{8} - \frac{\sin 4 x}{32} + C \end{array}$

(b) $\begin{array}{rcl} \frac{x}{8} - \frac{\sin 4 x}{32} + C \end{array}$

Step by step solution

Part (a) Step 1. Explanation

$\begin{array}{rcl} 鈭� \sin^{2} x \cos^{2} x d x & = & 鈭� (1 - \cos^{2} x) \cos^{2} x d x \\ = & 鈭� \cos^{2} x - \cos^{4} x d x \\ = & 鈭� \frac{1}{2} + \frac{\cos 2 x}{2} - \cos^{4} x d x \\ = & \frac{x}{2} + \frac{\sin 2 x}{4} - 鈭� \cos^{4} x d x \\ = & \frac{x}{2} + \frac{\sin 2 x}{4} - 鈭� \frac{\cos 2 x}{2} + \frac{\cos 4 x}{8} + \frac{3}{8} d x \\ = & \frac{x}{2} + \frac{\sin 2 x}{4} - \frac{\sin 2 x}{4} - \frac{\sin 4 x}{32} - \frac{3}{8} x + C \\ = & \frac{x}{8} - \frac{\sin 4 x}{32} + C \end{array}$

Part (b) Step 1. Explanation

$\begin{array}{rcl} 鈭� \sin^{2} x \cos^{2} x d x & = & 鈭� \frac{1}{4} \sin^{2} 2 x \\ = & \frac{1}{4} 鈭� \frac{1}{2} - \frac{\cos (4 x)}{2} d x \\ = & \frac{x}{8} - \frac{\sin 4 x}{32} + C \end{array}$

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Short Answer

Step by step solution

Part (a) Step 1. Explanation

Part (b) Step 1. Explanation

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Consider the integral 鈭�sin2xcos2xdx. We solved this integral in Example 2(b) by applying double-angle identities at the very beginning.(a) Solve this integral by applying the identity sin2x=1-cos2x(b) Solve this integral another way, by applying the identitysinxcosx=12sin2x

Short Answer

Step by step solution

Part (a) Step 1. Explanation

Part (b) Step 1. Explanation

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