Integration by substitution - I'm stuck

[jkristia]

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

 

[jkristia]

I know the answer is

$(sinθ - \frac{sin^3}{3}) + C$

and I can see how to differentiate this

$cosθ - \frac{3}{3}sin^2θcosθ$
= $(1- sin^2θ)cosθ$

But I can't see what I'm doing wrong when trying to go the opposite direction

 

[dextercioby]

Substitution or integration by parts ??

$$\int (1-\sin^2 x) \cos x ~ dx = \int (1-\sin^2 x) ~ d(\sin x) = ...$$

Do you see the substitution ?

 

[ehild]

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

 

[jkristia]

>>Substitute u=sinθ.

Ah yes of course.

∫(1 - u²) du

u - u³/3 + C

Thank you very much.