int3x-2cscxdx∫3x−2cscxdx
=int3xdx-2intcscxdx=∫3xdx−2∫cscxdx
To integrate the second integral re write and use the following substitution:
intcscxdx=intcscx(cscx+cotx)/(cscx+cotx)dx∫cscxdx=∫cscxcscx+cotxcscx+cotxdx
u=cscx+cotxu=cscx+cotx
-> du=-cotxcscx-csc^2xdx=-cscx(cscx+cotx)dx→du=−cotxcscx−csc2xdx=−cscx(cscx+cotx)dx
Now substitute into the integral:
intcscx(cscx+cotx)/(cscx+cotx)dx=-int(du)/u=-lnu+C_0∫cscxcscx+cotxcscx+cotxdx=−∫duu=−lnu+C0
Reverse the substitution:
-ln|u| +C_0 = -ln|csc(x)+cot(x)|+C_0−ln|u|+C0=−ln|csc(x)+cot(x)|+C0
So, returning to the original integral:
int3xdx-2intcscxdx=3/2x^2+2ln|csc(x)+cot(x)|+C∫3xdx−2∫cscxdx=32x2+2ln|csc(x)+cot(x)|+C
Now solve for CC.
f((5pi)/4)f(5π4)
=3/2((5pi)/4)^2+2ln|csc((5pi)/4)+cot((5pi)/4)|+C=0=32(5π4)2+2ln∣∣∣csc(5π4)+cot(5π4)∣∣∣+C=0
75/32pi^2+2ln(-sqrt(2)+1)+C7532π2+2ln(−√2+1)+C
->21.3691+C=0-> C=-21.3691 →21.3691+C=0→C=−21.3691
f(x) = 3/2x^2+2ln|csc(x)+cot(x)|-21.3691f(x)=32x2+2ln|csc(x)+cot(x)|−21.3691
