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# Utah State University - MATH 1220 - Class Notes - Week 2

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Utah State University - MATH 1220 - Class Notes - Week 2

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Colton Hill
January 15 2019
Math 1220-502
Homework Problems 3 1. Evaluate the following integrals. (a) ˆ sin 3 x cos x dx u = sin(x), dx = du cos(x) ˆ u 3 du u 4 4 + C sin 4 (x) 4 Solution: ˆ sin 3 x cos x dx = 1 4 sin 4 x + C (b) ˆ tan 5 (2x) sec 2 (2x) dx u = tan(2x), du = 2 sec 2 (2x) dx ˆ u 5 2 dx = u 6 12 + C tan 6 (2x) 12 + C Solution: ˆ tan 5 (2x) sec 2 (2x) dx = 1 12 tan 6 (2x) + C (c) ˆ sin 3 x dx ˆ sin(x)(1 − cos 2 (x)) dx u = cos(x), du = − sin(x) ˆ sin(x) ∗ (1 − cos 2 (x)) −sin(x) du ˆ −1 + cos 2 (x) du ˆ −1 + u 2 du −u + u 3 3 + C − cos(x) + cos 3 (x) 3 + C Solution: ˆ sin 3 x dx = 1 3 cos 3 x − cos x + C (d) ˆ cos 5 x dx ˆ cos(x)(1 − sin 2 ) 2 dx u = sin(x), du = cos(x) dx ˆ (1 − sin 2 (x)) 2 du ˆ (1 − u 2 ) 2 du 1
ˆ 1 − 2u 2 + u 4 + C u − 2 3 u 3 + 1 5 u 5 + C sin(x) − 2 3 sin 3 (x) + 1 5 sin 5 (x) + C Solution: ˆ cos 5 x dx = sin x − 2 3 sin 3 x + 1 5 sin 5 x + C (e) ˆ sin 3 x cos 3 x dx ˆ sin 3 x ∗ cosx(1 − sin 2 x) dx u = sinx, du = cosx dx ˆ sin 3 x(1 − sin 2 x) du ˆ u 3 − u 5 du 1 4 u 4 1 6 u 6 + C 1 4 sin 4 x − 1 6 sin 6 x + C Solution: ˆ sin 3 x cos 3 x dx = 1 4 sin 4 x − 1 6 sin 6 x + C (f) ˆ tan x sec 3 x dx u = sec(x) du = tan(x)sec(x) dx ˆ sec 2 (x) du ˆ sec 2 (x) du ˆ u 2 (x) du 1 3 u 3 + C 1 3 sec 3 (x) + C Solution: ˆ tan x sec 3 x dx = 1 3 sec 3 x + C (g) ˆ π 0 sin 2 x dx ˆ π 0 1 2 1 2 cos(2x) dx ˆ 1 2 dx − ˆ 1 2 cos(2x) dx x 2 1 4 sin(2x) + C| π
0
π 2 1 4 sin(2π) − [ 0 2 1 4 sin(0)] = π 2 Solution: ˆ π 0 sin 2 x dx = π 2 (h) ˆ π/8 0 sin 2 x cos 2 x dx ˆ π/8 0 1 2 (1 − cos(2x)) ∗ 1 2 (1 + cos(2x)) dx 1 4 ˆ 1 − cos 2 (2x) dx 1 4 ˆ 1 dx − 1 4 ˆ cos 2 (2x) dx 2

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##### Description: These notes cover the homework problems , with emphasis on rudimentary to advanced trig manipulation to facilitate integration
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