Integrate: cos2x/[cos^2 (x).sin^2 (x)]-cot(x)/2

[Tanishq Nandan]

Homework Statement

Integrate: cos2x/[cos^2 (x).sin^2 (x)]

Homework Equations

[/B]
▪cos2x=1-2sin^2 (x)
▪2sinxcosx=sin2x
▪1/sinx = cosecx
▪Integration of cosec^2 (ax+b)=[-cot(ax+b)]/a
▪Integration of sec^2 (x)=tanx

The Attempt at a Solution

I have attached my solution,but the answer is not matching with the correct answer (written in the last line).I wrote the given answer as well coz it may be a manipulation of my answer,which i can't see(doubtful,but not impossible).If anyone could just point out the line where I'm going wrong..

 

[BvU]

which i can't see(doubtful,but not impossible)

An easy check is to differentiate your result !

 

[BvU]

And you could convert $\cot 2x$ to $\ \ \displaystyle {\cot^2 x -1 \over 2\cot x}$

 

[Tanishq Nandan]

An easy check is to differentiate your result !

Of course,why didn't I think of that?
Thanks!

 

[Ray Vickson]

Homework Statement

Integrate: cos2x/[cos^2 (x).sin^2 (x)]

Homework Equations

[/B]
▪cos2x=1-2sin^2 (x)
▪2sinxcosx=sin2x
▪1/sinx = cosecx
▪Integration of cosec^2 (ax+b)=[-cot(ax+b)]/a
▪Integration of sec^2 (x)=tanx

The Attempt at a Solution

I have attached my solution,but the answer is not matching with the correct answer (written in the last line).I wrote the given answer as well coz it may be a manipulation of my answer,which i can't see(doubtful,but not impossible).If anyone could just point out the line where I'm going wrong..

You are developing a bad habit, which you should stop right away if you want to continue posting to PF. Most helpers will not look at images of handwritten solutions; I, for one, will not. You may be lucky to find somebody willing to help by looking at your images, but please do not keep doing it; the PF standard is to type out your work, and it really is not very difficult. For example, you can write $\int_a^b x/(x^2+a^2) \, dx$ in plain text as int{ x/(x^2+a^2) dx, x=a..b} (or as int_{x=a..b} {x/(x^2+a^2) dx}) and that is perfectly readable. Just be careful to use parentheses, so that $\frac{a + b}{c}$ is written as (a+b)/c, NOT as a + b/c (which means $a + \frac{b}{c}$).

Please try to reserve images for things like drawings, diagrams and/or data tables.