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

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Answer 1 · 2018-05-05T07:32:54

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

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

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#