What is #f(x) = int 3x-2cscx dx# if #f((5pi)/4) = 0 #?

Question

1691 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-11T21:37:28

#f(x) = 3/2x^2+2ln|csc(x)+cot(x)|-21.3691#

#int3x-2cscxdx#

#=int3xdx-2intcscxdx#

To integrate the second integral re write and use the following substitution:

#intcscxdx=intcscx(cscx+cotx)/(cscx+cotx)dx#

#u=cscx+cotx#
#-> du=-cotxcscx-csc^2xdx=-cscx(cscx+cotx)dx#

Now substitute into the integral:

#intcscx(cscx+cotx)/(cscx+cotx)dx=-int(du)/u=-lnu+C_0#

Reverse the substitution:

#-ln|u| +C_0 = -ln|csc(x)+cot(x)|+C_0#

So, returning to the original integral:

#int3xdx-2intcscxdx=3/2x^2+2ln|csc(x)+cot(x)|+C#

Now solve for #C#.

#f((5pi)/4)#
#=3/2((5pi)/4)^2+2ln|csc((5pi)/4)+cot((5pi)/4)|+C=0#

#75/32pi^2+2ln(-sqrt(2)+1)+C#

#->21.3691+C=0-> C=-21.3691 #

#f(x) = 3/2x^2+2ln|csc(x)+cot(x)|-21.3691#