I How Do You Solve the Integral of 1/(y+cos(x))^2?

[Needhelp62]

First part of the question was to work out the integral 1/(y+cos(x)) between x=0 and x=pi/2 by using the substitution t=tan(x/2).
Got this to be \frac{2}{\sqrt{y^2-1}}arctan(\sqrt{\frac{y-1}{y+1}})
The next question says HENCE find integral with the same limits of \frac{1}{(y+cos(x))^2}
Ive tried using by parts to maybe get something minus the first integral but i can't get it to work. This exam is known for being full of horrible questions so anything is possible here

 

[Ssnow]

Hi, but $y$ is a fixed number or is a function $y(x)$?

 

[member 587159]

https://en.wikipedia.org/wiki/Tange..._half-angle_substitution_in_integral_calculus

 

[Needhelp62]

Hi, but $y$ is a fixed number or is a function $y(x)$?

y is just a fixed number the question says the answer to the integral is a function of y

 

[Ssnow]

ok, in the risolution you must treat $y$ as a number and try with the substitution $t=\tan{\frac{x}{2}}$. You will obtain an integral that is the ratio of two polynomials in $t$ ...

 

[Ssnow]

remember that $\cos{x}=\frac{1-t^{2}}{1+t^{2}}$...

 

[Needhelp62]

remember that $\cos{x}=\frac{1-t^{2}}{1+t^{2}}$...

Thank you I get how the original integral i used now.I have two integrals with the first being twice the original and the second being
\frac{2t^2}{(y+1)+t^2(y-1)}

 

[Needhelp62]

Thank you I get how the original integral i used now.I have two integrals with the first being twice the original and the second being
\frac{2t^2}{(y+1)+t^2(y-1)}

think I've got it, if i call the original integral Q so
(2-\frac{2}{y+1})Q+\frac{2}{y+1}