Solution for Integral of sin2x/(1+cos^2x) using Substitution Method

[PhizKid]

Homework Statement

$\int \frac{sin2x}{1 + cos^{2}x} \textrm{ } dx$

Homework Equations

The Attempt at a Solution

$\int \frac{sin2x}{1 + cos^{2}x} \textrm{ } dx \\\\ \int \frac{2sinxcosx}{1 + cos^{2}x} \textrm{ } dx \\\\ u = cosx \\\\ -\int \frac{2u}{1 + u^{2}} \textrm{ } du \\\\ w = u^{2} \\\\ -\int \frac{1}{1 + w} \textrm{ } dw \\\\ q = 1 + w \\\\ -\int \frac{1}{q} \textrm{ } dq \\\\ -ln(q) = -ln(1 + w) = -ln(1 + u^{2}) = -ln(1 + cos^{2}x)$

I don't know what I did wrong

 

[Curious3141]

Homework Statement

$\int \frac{sin2x}{1 + cos^{2}x} \textrm{ } dx$

Homework Equations

The Attempt at a Solution

$\int \frac{sin2x}{1 + cos^{2}x} \textrm{ } dx \\\\ \int \frac{2sinxcosx}{1 + cos^{2}x} \textrm{ } dx \\\\ u = cosx \\\\ -\int \frac{2u}{1 + u^{2}} \textrm{ } du \\\\ w = u^{2} \\\\ -\int \frac{1}{1 + w} \textrm{ } dw \\\\ q = 1 + w \\\\ -\int \frac{1}{q} \textrm{ } dq \\\\ -ln(q) = -ln(1 + w) = -ln(1 + u^{2}) = -ln(1 + cos^{2}x)$

I don't know what I did wrong

Your answer is correct, except for the omission of the constant of integration. Which makes all the difference in the world, really.

Here's an easier way to approach the integral. Note that the denominator is, in fact: $\frac{1}{2}(3 + \cos 2x)$. Can you go from there?

You'll find that the final answer you get using that method is different from your answer by only a constant ($\ln 2$), which means the answers are equivalent.

 

[SammyS]

Homework Statement

$\int \frac{sin2x}{1 + cos^{2}x} \textrm{ } dx$

Homework Equations

The Attempt at a Solution

$\int \frac{sin2x}{1 + cos^{2}x} \textrm{ } dx \\\\ \int \frac{2sinxcosx}{1 + cos^{2}x} \textrm{ } dx \\\\ u = cosx \\\\ -\int \frac{2u}{1 + u^{2}} \textrm{ } du \\\\ w = u^{2} \\\\ -\int \frac{1}{1 + w} \textrm{ } dw \\\\ q = 1 + w \\\\ -\int \frac{1}{q} \textrm{ } dq \\\\ -ln(q) = -ln(1 + w) = -ln(1 + u^{2}) = -ln(1 + cos^{2}x)$

I don't know what I did wrong

What's the derivative of a constant?

 

[PhizKid]

Your answer is correct, except for the omission of the constant of integration. Which makes all the difference in the world, really.

Here's an easier way to approach the integral. Note that the denominator is, in fact: $\frac{1}{2}(3 + \cos 2x)$. Can you go from there?

You'll find that the final answer you get using that method is different from your answer by only a constant ($\ln 2$), which means the answers are equivalent.

I don't see where that denominator is. The solution says it's: -ln(cos(2x) + 3) + C

 

[SammyS]

I don't see where that denominator is. The solution says it's: -ln(cos(2x) + 3) + C

What is $\displaystyle \ \ -\ln(1 + \cos^{2}x)-(-\ln(\cos(2x) + 3))\ ?$

 

[PhizKid]

What is $\displaystyle \ \ -\ln(1 + \cos^{2}x)-(-\ln(\cos(2x) + 3))\ ?$

Isn't it $ln(\frac{1 + cos^2x}{cos(2x) + 3})$? But what does $ln(cos(2x) + 3)$ have to do with anything?

 

[Mute]

Isn't it $ln(\frac{1 + cos^2x}{cos(2x) + 3})$?

Try plotting that function.

 

[PhizKid]

Try plotting that function.

It looks like -ln(2)

 

[SammyS]

It looks like -ln(2)

I get ln(2) .

That's just a constant, so the two answers are equivalent.