Why Does Trig Substitution Yield Different Integral Results?

[Jrlinton]

Homework Statement

∫8cos^3(2θ)sin(2θ)dθ

Homework Equations

The Attempt at a Solution

rewrote the integral as:
8∫(1-sin^2(2θ))sin(2θ)cos(2θ)dθ
u substitution with u=sin(2θ) du=2cos(2θ)dθ
4∫(1-u^2)u du= 4∫u-u^3 du
4(u^2/2-u^4/4)+C
undo substitution and simplify
2sin^2(2θ)-sin^4(2θ)+C
The book gives an answer of:
-cos^4(2θ)+C

 

[Ray Vickson]

Homework Statement

∫8cos^3(2θ)sin(2θ)dθ

Homework Equations

The Attempt at a Solution

rewrote the integral as:
8∫(1-sin^2(2θ))sin(2θ)cos(2θ)dθ
u substitution with u=sin(2θ) du=2cos(2θ)dθ
4∫(1-u^2)u du= 4∫u-u^3 du
4(u^2/2-u^4/4)+C
undo substitution and simplify
2sin^2(2θ)-sin^4(2θ)+C
The book gives an answer of:
-cos^4(2θ)+C

Check your answer by differentiating it to see if you get back to $8 \cos^3 (2 \theta) \sin (2 \theta)$. Checking your answer by differentiation is something you should always do, every time.

 

[Jrlinton]

I do in fact get my original expression. Thanks.

 

[Mark44]

Homework Statement

∫8cos^3(2θ)sin(2θ)dθ

Homework Equations

The Attempt at a Solution

rewrote the integral as:
8∫(1-sin^2(2θ))sin(2θ)cos(2θ)dθ
u substitution with u=sin(2θ) du=2cos(2θ)dθ

There's no need to rewrite the integral. In the original integral, use ordinary substitution with $u = \cos(2\theta)$, and $du = -2\sin(2\theta)d\theta$. Using this substitution leads directly to the book's answer.

4∫(1-u^2)u du= 4∫u-u^3 du
4(u^2/2-u^4/4)+C
undo substitution and simplify
2sin^2(2θ)-sin^4(2θ)+C
The book gives an answer of:
-cos^4(2θ)+C

I second Ray's advice to always check your answer by differentiation.