Integration

[meager echo]

Integration ( \int_0^\infty \frac{x}{(1+x)(1+x^2)} , dx ).

So i put x=tan theta and i got sin2theta/(1+sintheta)o

[tranquil timber]

Andy

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[ornate ruin]

There's no need for any substitution. Just use partial fractions.

[meager echo]

I have solved out by partial fractions

But i wanted to solve in this way too

And i got π/4

[ornate ruin]

Well, if that's the same answer you got as before, then that's probably correct.

[meager echo]

What?

I got my answer by partial fraction integration but I want to do it by substitution method also@ornate ruin

[ornate ruin]

Ah, I see.
Well, if you just take x = tan(t), I don't think you'd get what you wrote. Can you show your process?

[meager echo]

What process?

[ornate ruin]

The process of your substitution.

[meager echo]

just put x=tan theta and i got sin(2theta)/(1+sin theta)

[ornate ruin]

No, that's not what you get.

That's why I'm asking.

[meager echo]

Ohh correct

Slightly mistake

twn theta/sec^2theta×(1+tan theta)

[ornate ruin]

Well, I was talking more about the process.

We have:
x = tan(θ), θ = arctan(x)
dθ = dx/(1 + x^2)
So:
x dx/((1 + x)(1 + x^2)) = tan(θ)dθ/(1 + tan(θ)) = sin(θ)dθ/(cos(θ) + sin(θ))
And the new limits then are from 0 to π/2.

As for how to get the result... Well, I do have an idea.

Take that integral as I, then perform a substitution u = π/2 - θ.

[meager echo]

I didn't understand how u=π/2-x helps?@ornate ruin

[ornate ruin]

Well, first of all, what do you get?

[meager echo]

Where? We have this form

Wait

[ornate ruin]

Ah! Sorry. I miswrote here.

One sec...

Ok, corrected.

What I mean is that you should do it after the first substitution.

[meager echo]

@ornate ruin

[ornate ruin]

You didn't simplify correctly. tan(θ)dθ/(1 + tan(θ)) = sin(θ)dθ/(cos(θ) + sin(θ)), not just sin(θ)dθ/(1 + sin(θ)).

[meager echo]

So we get same thing after this substitution

[ornate ruin]

Why do you get an extra cosine in the numerator? It shouldn't be there.

[ornate ruin]

Again, as I said, tan(θ)dθ/(1 + tan(θ)) = sin(θ)dθ/(cos(θ) + sin(θ)).

[meager echo]

Fuck

[ornate ruin]

Yeah, nice!

[meager echo]

Two mistakes

And reached at distance

[ornate ruin]

So, the point is that by doing that substitution the sine in the numerator changes into cosine, and nothing else changes.

So, you can do I + I, which turns out to be a constant.

[meager echo]

My next doubt

Just give hints

Integration of sinx/(cos(x-a))

[ornate ruin]

Indefinite?

If so, I recommend starting with u = x - a.

[meager echo]

Yeah

[meager echo]

+close

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