A possible way of computing this integral int_(pi/4)^(pi/2)cscx - sinx dx, is by simplifying to give; int_(pi/4)^(pi/2)cosxcotx dx But a more interesting method is via individually finding intcscx dx
let u = cotx
then du = -csc^2 x dx
Via quotient rule;
cotx = cosx/sinx
hence d/dx(cotx) = ((sinx)(-sinx)-(cosx)(cosx))/sin^2 x
Hence = -csc^2 x
Hence (-du)/cscx = cscx dx
Hence intcscx dx becomes -int (du)/cscx
Considering 1 + cot^2 x = csc^2 x
Hence if u = cotx then cscx = (1+u^2)^(1/2)
Hence -int (du)/cscx becomes -int (du)/(1+u^2)^(1/2)
Then make a new substitution of u = sinhtheta
Hence du = coshtheta d theta
Now by cosindering cosh^2 theta - sinh^2 theta = 1
-int (du)/(1+u^2)^(1/2) becomes -int d theta
= -theta + c
As u = sinh theta then theta = arcsinh(u)
Hence -arcsinh (u) + c
Hence as u = cottheta
Hence int csc xdx = c - sinh^-1(cotx)
Hence int csc x- sinxdx = cosx - sinh^-1(cotx) +c
Hence int_(pi/4)^(pi/2)cscx - sinx dx = | cosx - sinh^-1(cotx) | from pi/4 to pi/2
So hence by using sinh^-1(x) = ln( x + (1+x^2)^(1/2)) , We can simply evaluate to give;
ln(1+2^(1/2)) -2^(-1/2)
