# Spring 2013 Homework 2 Solution - 7.3.16 We make the...

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Unformatted text preview: 7.3.16. We make the substitution x = 1 3 sec θ , keeping in mind that ≤ θ < π 2 or π ≤ θ < 3 π 2 . Then dx = 1 3 sec θ tan θdθ . To change the bounds, we make use of the fact that sec θ = (cos θ )- 1 . For x = 2 3 we have 2 3 = 1 3 sec θ 2 = 1 cos θ cos θ = 1 2 and ≤ θ < π 2 implies θ = π 3 . Similarly, for x = √ 2 3 we obtain θ = π 4 . Continuing with the substitution, we have ˆ 2 / 3 √ 2 / 3 dx x 5 √ 9 x 2- 1 = ˆ π/ 3 π/ 4 1 3 sec θ tan θdθ ( 1 3 sec θ ) 5 q 9 ( 1 3 sec θ ) 2- 1 = ˆ π/ 3 π/ 4 tan θdθ ( 1 3 sec θ ) 4 √ tan 2 θ . Next, we note that √ tan 2 θ = | tan θ | . And ≤ θ < π 2 or π ≤ θ < 3 π 2 so that tan θ ≥ and hence | tan θ | = tan θ. ˆ π/ 3 π/ 4 tan θdθ ( 1 3 sec θ ) 4 √ tan 2 θ = ˆ π/ 3 π/ 4 tan θdθ ( 1 3 sec θ ) 4 tan θ = 3 4 ˆ π/ 3 π/ 4 (cos θ ) 4 dθ Now (cos θ ) 4 = ( cos 2 θ ) 2 = 1 + cos 2 θ 2 2 = 1 + 2 cos 2 θ + cos 2 2 θ 4 = 1 + 2 cos 2 θ + 1+cos 4 θ 2 4 !...
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