Solve this integral involving a fraction with a trig function

[Rhdjfgjgj]

Homework Statement

Hello guys. Please look at the following problem .(image).
The first line is the the problem statement. Later is my attempt to solve it. I have done the working as per the standard techniques taught by my sir. But I end up at the wrong answer. What may I have done wrong. the actual answer is 2pi/√3. Please point out the steps which are wrong or give a correct approach.

Relevant Equations

Properties of definite integrals and standard integral formulae.

 

[Mark44]

Your second line looks wrong to me. It appears that you replaced $\sin(2x)$ with the identity for $\tan(2x)$ except that the denominator should be $1 - \tan^2(x)$ not $1 + \tan^2(x)$. In any case, that's not the correct identity for $\sin(2x)$.

It's hard for me to read much of what you wrote after that, and this is one of the reasons we discourage photos of hand-written work.

 

[Rhdjfgjgj]

No there's this formula that goes like
Sin2x=2tanx/(1+tan^2x)
It's a standard trigonometric formula.

 

[Rhdjfgjgj]

Im sorry about the picture.

 

[Mark44]

No there's this formula that goes like
Sin2x=2tanx/(1+tan^2x)
It's a standard trigonometric formula.

OK, you're right. I can't read what you have from the third line on...

 

[haruspex]

Homework Statement: Hello guys. Please look at the following problem .(image).
The first line is the the problem statement. Later is my attempt to solve it. I have done the working as per the standard techniques taught by my sir. But I end up at the wrong answer. What may I have done wrong. the actual answer is 2pi/√3. Please point out the steps which are wrong or give a correct approach.
Relevant Equations: Properties of definite integrals and standard integral formulae.

View attachment 336195

I follow you down to the third line, equivalent to $\frac 12\int\frac{dt}{1+t+t^2}$, but don't see how you got the next line.

 

[Rhdjfgjgj]

I completed the square
1+t+t^2 =1+2(1/2)t+(1/2)^2+3/4

 

[haruspex]

I completed the square
1+t+t^2 =1+2(1/2)t+(1/2)^2+3/4

Ah, I read the t in the denominator as a 4.
I would say your problem is the behaviour of tan as the angle goes from 0 to 2π.
Try breaking up the range, dealing with each improper integral carefully and watching how you evaluate arctan. Remember, the integrand you started with is everywhere positive.

 

[Rhdjfgjgj]

Ah, I read the t in the denominator as a 4.
I would say your problem is the behaviour of tan as the angle goes from 0 to 2π.
Try breaking up the range, dealing with each improper integral carefully and watching how you evaluate arctan. Remember, the integrand you started with is everywhere positive.

The end result of improper integral itself gave the hint that I had to break the limit.
But where, I would be glad if you could help me out on this

 

[haruspex]

The end result of improper integral itself gave the hint that I had to break the limit.
But where, I would be glad if you could help me out on this

Break it at each point where the argument of the arctan is unbounded, or where t is unbounded, which is the same.

 

[SammyS]

No there's this formula that goes like
Sin2x=2tanx/(1+tan^2x)
It's a standard trigonometric formula.

It is the tangent half-angle formula for sine, in disguise. I wouldn't call it standard.

This entire thread would be more accessible if more LaTeX was used.

 

[haruspex]

It is the tangent half-angle formula for sine, in disguise. I wouldn't call it standard.

Oh. I would.