Evaluating an Indefinite Integral using Substitution

[Drakkith]

Homework Statement

Evaluate the Integral:
∫sin2x dx/(1+cos²x)

Homework Equations

The Attempt at a Solution

I first broke the numerator up:
∫2sinxcosx dx /(1+cos²x)
2∫sinxcosx dx /(1+cos²x)

Then I let u = cosx so that du = -sinx dx
-2∫u du/(1+u²)

And now I'm stuck. I thought about turning 1+u² into tan^-1u + c, but I still have another u to deal with in the numerator.

 

[SammyS]

Homework Statement

Evaluate the Integral:
∫sin2x dx/(1+cos²x)

Homework Equations

The Attempt at a Solution

I first broke the numerator up:
∫2sinxcosx dx /(1+cos²x)
2∫sinxcosx dx /(1+cos²x)

Then I let u = cosx so that du = -sinx dx
-2∫u du/(1+u²)

And now I'm stuck. I thought about turning 1+u² into tan^-1u + c, but I still have another u to deal with in the numerator.

Try a further substitution.

What is the derivative of 1 + u² ?

 

[Drakkith]

Try a further substitution.

What is the derivative of 1 + u² ?

Okay. So if I let y = 1+u², then dy = 2u du, or dy/2 = u du

That gives me -2/2∫y^-1 dy
-∫y^-1 dy

Integrating:
-ln|y| + c

Since y = 1+u² that becomes:
-ln|1+u²|+ c

And since u = cosx:
-ln |1 + cos²x| + c

 

[ahmed markhoos]

Homework Statement

Evaluate the Integral:
∫sin2x dx/(1+cos²x)

Homework Equations

The Attempt at a Solution

I first broke the numerator up:
∫2sinxcosx dx /(1+cos²x)
2∫sinxcosx dx /(1+cos²x)

Then I let u = cosx so that du = -sinx dx
-2∫u du/(1+u²)

And now I'm stuck. I thought about turning 1+u² into tan^-1u + c, but I still have another u to deal with in the numerator.

hint: differentiation of $ln(x) = \frac{1}{x}$

 

[SammyS]

Okay. So if I let y = 1+u², then dy = 2u du, or dy/2 = u du

That gives me -2/2∫y^-1 dy
-∫y^-1 dy

Integrating:
-ln|y| + c

Since y = 1+u² that becomes:
-ln|1+u²|+ c

And since u = cosx:
-ln |1 + cos²x| + c

Right.

So looking back at your two substitutions, letting u = 1 + cos²(x) would have worked very nicely.

 

[epenguin]

Er, before you do any calculus can't you very significantly simplify the thing you are asked to integrate?

 

[Drakkith]

Right.

So looking back at your two substitutions, letting u = 1 + cos²(x) would have worked very nicely.

How so? I'm not getting anything that looks useful.

u = 1 + cos²x, du = -2sinx dx
That still leaves cosx in the numerator.

Er, before you do any calculus can't you very significantly simplify the thing you are asked to integrate?

I'm open to suggestions.

 

[SammyS]

How so? I'm not getting anything that looks useful.

u = 1 + cos²x, du = -2sinx dx
That still leaves cosx in the numerator.

I'm open to suggestions.

Check your differentiation.

 

[Drakkith]

Check your differentiation.

Oh. Apparently it's -2sinxcosx. Yeah that would work nicely.

It's been a while since I did any differentiation and I appear to have forgotten some important details about how to use the chain rule.

 

[epenguin]

Not only?

 

[Drakkith]

Not only?

Huh?

 

[epenguin]

Huh?

OK my late night mistake..

 

[Drakkith]

OK my late night mistake..

No worries!

Anyways, thanks for the help all! Sammy, I think I'll have to spend a little time going of differentiation rules again...

 

[SammyS]

No worries!

Anyways, thanks for the help all! Sammy, I think I'll have to spend a little time going of differentiation rules again...

Sure. Practice helps.

It was interesting that you got the part of the chain rule that most people miss and missed the part they get.

 

[epenguin]

Oh. Apparently it's -2sinxcosx. Yeah that would work nicely.

It's been a while since I did any differentiation and I appear to have forgotten some important details about how to use the chain rule.

Conclusion?

 

[SammyS]

Conclusion?

This was the problem Drakkith was asking about:

Homework Statement

Evaluate the Integral:
∫sin2x dx/(1+cos2x)

This was his final result.

...
And since u = cosx:
-ln |1 + cos2x| + c

After that I suggested a single substitution to do the job, which he indicates he now adequately understands.

 

[epenguin]

Ah yes, I had not read carefully enough, I just read 'substitution' and thought oof, that is a bit heavy. In fact in #1 he was only missing that the numerator is (apart from numerical factor) is the derivative of the denominator. I had just mentally written
$$2 \int \frac{cos x sin x}{1 + cos^2 x} dx$$
$$. = - 2 \int \frac{cos x }{1 + cos^2 x} d (cos x)$$
$$. =-. \int \frac{d (cos^2 x)}{1 + cos^2 x}.$$
$$. = - \int \frac{d (1 + cos^2 x)}{1 + cos^2 x}.$$
$$. = - ln (1 + cos^2 x) + C$$
which you can also write
$$. = - ln K (1 + cos^2 x)$$

which is just the some thing as Drakkith, just slightly differenti setting out.