What is f(x) = int xcotx^2 dx if f((5pi)/4) = 0 ?

1 Answer
May 5, 2018

I have tried , but it is complicated for x^2=((5pi)/4)^2.

f(x)=1/2ln|sinx^2|+c,where, c~~0.6623

Explanation:

Here,

f(x)=I=intxcotx^2dx=intcotx^2*xdx

Let, x^2=u=>2xdx=du=>xdx=1/2du

So,

I=intcotu*1/2du

=1/2ln|sinu|+c,where,u=x^2

=>f(x)=1/2ln|sinx^2|+c...to(A)

Given that,

f((5pi)/4)=0

=>1/2ln|sin((5pi)/4)^2|+c=0

=>2c=-ln|sin((25pi^2)/16)|

=>2c=ln|(1/(sin((25pi^2)/16)))|

=>c=1/2ln|(1/(sin((25pi^2)/16)))|...to(1)

Now,

(25xxpi^2)/16 ~~15.42....

sin((25pi^2)/16)~~0.2659... in[-1,1]

1/(sin((25pi^2)/16))~~3.7606...

ln|1/(sin((25pi^2)/16))| ~~1.3245...

1/2*ln|1/(sin((25pi^2)/16))| ~~0.6623

Hence, from (1)

c ~~ 0.6623

Thus,

f(x)=1/2ln|sinx^2|+c,where, c~~0.6623