Slide #1.

Integration by Parts & Trig Functions Chapter 7.1 March 22, 2007
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Slide #2.

Integration by Parts Evaluate:  y 3 5 ∫y e ∫udv =uv−∫vdu dy Choose u (to differentiate (“du”)) dv (to integrate (“v”)) Chapter 7.1 March 22, 2007
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Slide #3.

Integration by Parts ∫udv =uv−∫vdu 0 Evaluate:  ∫(2x + 6)e −x dx −1 Choose u (to differentiate (“du”)) dv (to integrate (“v”))
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Slide #4.

Integration by Parts Integrate: ∫cos (ln(x))dx ∫udv =uv−∫vdu
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Slide #5.

3 Integrate:  ∫x −1 x + 1 dx Choose u (to differentiate (“du”)) dv (to integrate (“v”))
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Slide #6.

3 Integrate:  ∫x −1 x + 1 dx You can also use u-substitution!
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Slide #7.

Integration by Parts ∫udv =uv−∫vdu Examples: 3 x2 ∫x e dx ∫tan x ln(cos x)dx x x e cos(e )dx ∫ x e ∫ cos(4 x)dx
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Slide #8.

Trig Functions: (also on Day 10) Recall that when a trig integral has ODD powers, we save one term for the “du” and replace the remaining even powers with: sin 2 x =1 −cos2 x cos 2 x =1 −sin2 x And when a trig integral has EVEN powers, we use the following trig identities to integrate. 1 sin 2 x = (1 −cos(2 x)) 2 1 cos 2 x = (1 + cos(2 x)) 2
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Slide #9.

Integrate: ⎛ t ⎞ 2 ⎛ t⎛ cot ∫ ⎜⎝ 2 ⎟⎠ sin ⎛⎛ 2 ⎛⎛ dt 2 Solution: t ⎛ t⎛ ⎛ t⎛ + cos⎛ ⎛ sin ⎛ ⎛ ⎛ 2⎛ ⎛ 2⎛ 2
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Slide #10.

Or you can use Integration by Parts ⎛t ⎞ ∫cos ⎜⎝ 2 ⎟⎠ dx 2 ∫udv =uv−∫vdu
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