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Integration by substitution - I'm stuck

  • Thread starter jkristia
  • Start date
  • #1
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θ

Homework Statement



I'm trying to do an integration by substitution, but I'm completely stuck at the moment

∫(1-sin2θ)cosθ dθ

Homework Equations



∫u dv = uv - ∫v du

The Attempt at a Solution



u = 1 - sin2θ
dv = cosθ dθ

du = -2sinθcosθ or -sin(2θ)
v = sin

I found du as the derivative of (1 - sin2)
= 0 - 2(sin)(cos) = -2sinθcosθ

Now when I insert I get

(1 - sin2θ)(sinθ) - ∫(sinθ)(-2sinθcosθ)
sinθ-sin3θ - ∫-2 sin2θcosθ
sinθ-sin3θ - (-2) ∫sin2θcosθ
sinθ-sin3θ - (-2) ∫(1-cos2θ)cosθ
sinθ-sin3θ - (-2) ∫cosθ-cos3θ

I think I'm on the wrong track here.
Where did I go wrong? - any help is much appreciated.

Thanks
 

Answers and Replies

  • #2
54
0
I know the answer is

[itex](sinθ - \frac{sin^3}{3}) + C[/itex]

and I can see how to differentiate this

[itex]cosθ - \frac{3}{3}sin^2θcosθ[/itex]
= [itex](1- sin^2θ)cosθ[/itex]

But I can't see what I'm doing wrong when trying to go the opposite direction
 
  • #3
dextercioby
Science Advisor
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Insights Author
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Substitution or integration by parts ??

[tex] \int (1-\sin^2 x) \cos x ~ dx = \int (1-\sin^2 x) ~ d(\sin x) = ... [/tex]

Do you see the substitution ?
 
Last edited:
  • #4
ehild
Homework Helper
15,477
1,854
θ

Homework Statement



I'm trying to do an integration by substitution, but I'm completely stuck at the moment

∫(1-sin2θ)cosθ dθ

Homework Equations



∫u dv = uv - ∫v du
You tried to do integration by parts, instead of substitution. Substitute u=sinθ.

ehild
 
  • #5
54
0
>>Substitute u=sinθ.

Ah yes of course.

∫(1 - u2) du

u - u3/3 + C

Thank you very much.
 

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