Determine the value of a constant C such that:

∫₀^pi/3(sin y/cos²y)dy = ∫₀^C(√(z+C) - √z)^-1dz

Note: The range of the first integral reads 'pi/3'.

lets start by evaluating the left integral

for simplicities sake all the following integrals are in terms of y and the range is from 0 to pi/3

integral(sin y/cos^2(y))

now we know that cos^2(y)=1-sin^2(y) so if we use the substitution of x=sin(y)  then we get this new integral with the new range is from 0 to sqrt(3)/2

integral(x/(1-x^2))

now if we use the substitution u=1-x^2 we have du=-2x dx thus we get yet another integral with range from 1 to 1-(sqrt(3)/2)

integral(-0.5*u^-1)

-0.5*integral(u^-1)

and this can easily be evaluated out to be equal to 1

now we need to evaluate the second integral.  This at first looks rather intimidating but if you multiply both the numerator and denominator by sqrt(z+c)+sqrt(z) then the denominator simplifies to c and thus we are left with a much simpler integral

(1/c)*integral(sqrt(z+c)+sqrt(z))

now this can easily be evaluated out to 4*sqrt(2c)/3

setting this equal to 1 and solving for c we get

c=9/32

Posted by Daniel on 2007-10-24 18:18:26