Evaluate the following integral:

[Synchronised]

Homework Statement

Evaluate the following integral using the substitution u=\frac{1-x}{1+x}

Homework Equations

\int_{0}^{1}\frac{ln(1+x)}{1+x^{2}}dx

The Attempt at a Solution

I tried using the given substitution but couldn't do the integral but I did it using u=tan^{-1}x and got the answer = \frac{\pi ln2}{8} so how can I do it with the given substitution?

 

[SammyS]

Homework Statement

Evaluate the following integral using the substitution u=\frac{1-x}{1+x}

Homework Equations

\int_{0}^{1}\frac{ln(1+x)}{1+x^{2}}dx

The Attempt at a Solution

I tried using the given substitution but couldn't do the integral but I did it using u=tan^{-1}x and got the answer = \frac{\pi ln2}{8} very easily so how can I do it with the given substitution?

Using your substitution, u=tan^-1(x), what did you do with the ln(1+x) ?I suspect that there is a typo in that integral . The suggested substitution, \displaystyle \ \ u=\frac{1-x}{1+x},\ \ works quite well to evaluate the integral, \displaystyle \ \ \int_{0}^{1}\frac{\ln(1+x)}{(1+x)^{2}}\,dx\ .

 

[Synchronised]

Using your substitution, u=tan^-1(x), what did you do with the ln(1+x) ?I suspect that there is a typo in that integral . The suggested substitution, \displaystyle \ \ u=\frac{1-x}{1+x},\ \ works quite well to evaluate the integral, \displaystyle \ \ \int_{0}^{1}\frac{\ln(1+x)}{(1+x)^{2}}\,dx\ .

No, I am sure there is no typo in the integral.
The way I did it was: Let u=tan^{-1}x\therefore dx=(1+x^{2})du\therefore\int_{0}^{\frac{\pi }{4}}ln(1+tanu)du=\int_{0}^{\frac{\pi }{4}}ln(1+tanx)dx=\int_{0}^{\frac{\pi }{4}}ln(1+tan(\frac{\pi }{4}-x))dx=
\int_{0}^{\frac{\pi }{4}}ln(1+\frac{tan\frac{\pi }{4}-tanx}{1+tan\frac{\pi }{4}tanx})dx=\int_{0}^{\frac{\pi }{4}}ln(1+\frac{1-tanx}{1+tanx})dx=
\int_{0}^{\frac{\pi }{4}}ln(\frac{2}{1+tanx})dx=\int_{0}^{\frac{\pi }{4}}ln(2)dx-\int_{0}^{\frac{\pi }{4}}ln(1+tanx)dx
\therefore 2\int_{0}^{\frac{\pi }{4}}ln(1+tanx)dx=\int_{0}^{\frac{\pi }{4}}ln(2)dx
\therefore \int_{0}^{\frac{\pi }{4}}ln(1+tanx)dx=\frac{\pi ln2}{8}

 

[Millennial]

The integration shown above is correct. I also verified it using Mathematica just in case.

The given substitution does not seem very well to me. Your substitution, to be honest, would be my first try.

 

[Synchronised]

But the question says use the given substitution so I don't think I can get full marks if any by using a different substitution in the exam. Maybe the given substitution has something to do with \int_{0}^{\frac{\pi }{4}}ln(1+\frac{1-tanx}{1+tanx})dx ? they look similar.

 

[Synchronised]

You should break that long line further.

That looks better :)

 

[Dick]

But the question says use the given substitution so I don't think I can get full marks if any by using a different substitution in the exam. Maybe the given substitution has something to do with \int_{0}^{\frac{\pi }{4}}ln(1+\frac{1-tanx}{1+tanx})dx ? they look similar.

To use the given substitution, just solve u=(1-x)/(1+x) for x. Then substitute that into ln(1+x). Yes, it does look pretty similar.

 

[Synchronised]

To use the given substitution, just solve u=(1-x)/(1+x) for x. Then substitute that into ln(1+x). Yes, it does look pretty similar.

Ok, thank you :)

Letu=tan^{-1}x\therefore dx=(1+x^{2})du\therefore\int_{0}^{\frac{\pi }{4}}ln(1+x)du
Let u=\frac{1-x}{1+x}\therefore x=\frac{1-u}{1+u}\therefore \int_{0}^{\frac{\pi }{4}}ln(1+\frac{1-u}{1+u})du...

 

[SammyS]

The suggested substitution works perfectly well, when used right at the outset.

Your method has a curious step ...

\displaystyle \int_{0}^{\pi /4}\ln(1+tanx)dx=\int_{0}^{\pi /4}\ln(1+tan(\frac{\pi }{4}-x))dx

Of course, \displaystyle \ \tan(x)\ne\tan(\pi/4-x)\,,\ so what gives? This step can be shown to be valid by using the substitution, t = π/4 - x, then changing to variable of integration back to x.

 

[Dick]

The suggested substitution works perfectly well, when used right at the outset.

Your method has a curious step ...

\displaystyle \int_{0}^{\pi /4}\ln(1+tanx)dx=\int_{0}^{\pi /4}\ln(1+tan(\frac{\pi }{4}-x))dx

Of course, \displaystyle \ \tan(x)\ne\tan(\pi/4-x)\,,\ so what gives? This step can be shown to be valid by using the substitution, t = π/4 - x, then changing to variable of integration back to x.

It looks like the sort of trick you might pull to evaluate a definite integral when there is no easy antiderivative. Integrate backwards and apply a trig identity. It's just being clever. Too clever. The given substitution gives you an indefinite integral without the need to resort to tricks like that.

 

[Synchronised]

The suggested substitution works perfectly well, when used right at the outset.

Your method has a curious step ...

\displaystyle \int_{0}^{\pi /4}\ln(1+tanx)dx=\int_{0}^{\pi /4}\ln(1+tan(\frac{\pi }{4}-x))dx

Of course, \displaystyle \ \tan(x)\ne\tan(\pi/4-x)\,,\ so what gives? This step can be shown to be valid by using the substitution, t = π/4 - x, then changing to variable of integration back to x.

It's just a method to evaluate definite integrals. \int_{0}^{a}f(x)dx=\int_{0}^{a}f(a-x)dx to prove it you just need to use a dummy variable let u=a-x. therefore x=a-u and du=-dx \therefore \int_{0}^{a}f(x)dx=\int_{a}^{0}f(a-u)(-du)=\int_{0}^{a}f(a-u)(du)=\int_{0}^{a}f(a-x)(dx)

 

[haruspex]

u=\frac{1-x}{1+x}\therefore x=\frac{1-u}{1+u}\therefore \int_{0}^{\frac{\pi }{4}}ln(1+\frac{1-u}{1+u})du...

I don't get that. I get a factor 1/u in the integrand.
dx = -2(1+u)^-2du; (1+x²)^-1 = (1+u)²/(4u); (1+x²)^-1dx = -du/(2u).

 

[Synchronised]

I don't get that. I get a factor 1/u in the integrand.
dx = -2(1+u)^-2du; (1+x²)^-1 = (1+u)²/(4u); (1+x²)^-1dx = -du/(2u).

yeah you're right, i still can't evaluate the integral using the given sub from the start...

 

[SammyS]

Ok, thank you :)

Let\ u=tan^{-1}x\therefore dx=(1+x^{2})du\therefore\int_{0}^{\frac{\pi }{4}}ln(1+x)du
Let\ u=\frac{1-x}{1+x}\therefore x=\frac{1-u}{1+u}\therefore \int_{0}^{\frac{\pi }{4}}ln(1+\frac{1-u}{1+u})du...

I overlooked the above post earlier.

I hope you didn't mean that your original integral was equivalent to \displaystyle \ \int_{0}^{\frac{\pi }{4}}ln(1+\frac{1-u}{1+u})du\ .

Those are two different substitutions.

..., i still can't evaluate the integral using the given sub from the start...

Using the suggested substitution:

\displaystyle x=\frac{1-u}{1+u}=\frac{2}{1+u}-1\ \ so that \displaystyle \ \ x+1=\frac{2}{1+u}\ .

Also, \displaystyle \ \ 1+x^2=1+\left(\frac{1-u}{1+u}\right)^2=\frac{1+u^2}{(1+u)^2}\ .

Put that together with what haruspex gave you for dx & see where that all leads you.

Added in Edit:

There is a mistake in my last line. It should be:

\displaystyle \ \ 1+x^2=1+\left(\frac{1-u}{1+u}\right)^2=2\frac{1+u^2}{(1+u)^2}\ .

 

[Synchronised]

I overlooked the above post earlier.

I hope you didn't mean that your original integral was equivalent to \displaystyle \ \int_{0}^{\frac{\pi }{4}}ln(1+\frac{1-u}{1+u})du\ .

Those are two different substitutions.
Using the suggested substitution:

\displaystyle x=\frac{1-u}{1+u}=\frac{2}{1+u}-1\ \ so that \displaystyle \ \ x+1=\frac{2}{1+u}\ .

Also, \displaystyle \ \ 1+x^2=1+\left(\frac{1-u}{1+u}\right)^2=\frac{1+u^2}{(1+u)^2}\ .

Put that together with what haruspex gave you for dx & see where that all leads you.

I get 2\int_{0}^{1}\frac{ln(\frac{2}{1+u})}{1+u^{2}}du
So where do I go from here?

 

[haruspex]

I get -2\int_{0}^{1}\frac{ln(\frac{2}{1+u})}{1+u^{2}}du
So where do I go from here?

You can write ln(2/(1+u)) a little differently. Then you'll see that one part of the integral looks very like what you started with, but with a constant multiplier.

 

[Synchronised]

You can write ln(2/(1+u)) a little differently. Then you'll see that one part of the integral looks very like what you started with, but with a constant multiplier.

I almost got it, I get \int_{0}^{1}\frac{ln(1+x)}{1+x^{2}}dx=\frac{2ln2}{3}\int_{0}^{1}\frac{dx}{1+x^{2}} =\frac{\pi ln2}{6} so I made a mistake somewhere :(

 

[haruspex]

I think you got a sign wrong, leading to the divisor of 3.

 

[Synchronised]

what should the divisor be? I keep getting 3.

Edit: It should be 4 but I really can't see the mistake, I checked the signs and everything...

 

[SammyS]

I get 2\int_{0}^{1}\frac{ln(\frac{2}{1+u})}{1+u^{2}}du
So where do I go from here?

I edited post #14, so divide the above by 2.

That should give you the same answer you originally got.

 

[Synchronised]

Finally got it! Thank you so much guys :)