# Integration by substitution - I'm stuck

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

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$(sinθ - \frac{sin^3}{3}) + C$

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
Homework Helper
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 ?

Last edited:
ehild
Homework Helper
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

>>Substitute u=sinθ.

Ah yes of course.

∫(1 - u2) du

u - u3/3 + C

Thank you very much.