Answered: Evaluate the integrals ∫sin 2θ - cos…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter6: The Trigonometric Functions

Section6.4: Values Of The Trigonometric Functions

Problem 38E

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Question

Evaluate the integrals ∫sin 2θ - cos 2θ/(sin 2θ + cos 2θ)3 dθ

Expert Solution

Step 1

We have to evaluate the integrals:

$\int \frac{(\sin 2 θ - \cos 2 θ)}{{(\sin 2 θ + \cos 2 θ)}^{3}} d θ$

This integration will be solve d by substitution method since derivatives of denominator is in numerator.

So let $t = \sin 2 θ + \cos 2 θ$

Differentiating both sides with respect to $' θ'$ , we get

$\begin{array}{rcl} t & = & \sin 2 θ + \cos 2 θ \\ \frac{d t}{d θ} & = & \frac{d \sin 2 θ}{d θ} + \frac{d \cos 2 θ}{d θ} \\ = & \cos 2 θ \frac{d 2 θ}{d θ} - \sin 2 θ \frac{d 2 θ}{d θ} \\ = & \cos 2 θ (2) - \sin 2 θ (2) \\ \frac{d t}{d θ} & = & - 2 (\sin 2 θ - \cos 2 θ) \\ \frac{- d t}{2} & = & (\sin 2 θ - \cos 2 θ) d θ \end{array}$

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