Hello, L_U_K_E!
Just one more step . . .
Got it?
∫(sinx + secx) / tanx dx
∫(sinx/tanx)dx + ∫(secx/tanx)dx
∫(cosx)dx + ∫(1/sinx)dx
sinx + ... not sure where to go from here. tried to go ∫(sinx/sinx^2)
1/2∫(sinx/1-cos2x)
u=1-cos2x du=2sinxdx dx=du/2sinx
1/2∫(sinx/2)(du/2sinx)
1/4∫(1/u)du
so... sinx + 1/4(ln|1-cos2x|)
but the back of the book says the answer is sinx + ln | cscx-cotx |
any ideas??? help!!!!!!!!!!!!!!!!!!!!!!!!!!!
Or use the usual Weiestrass substitution.
hay i have the solution after simplfying u will reach to(let S is integral) S(cosx+cscx)dx=
S(cosx)dx+S(cscx)dx=
so the proplem to integrate csx look multiply by csx up and down u will reach to csx^2x/cscx remove the down one and write root of(1+cotan^2x) and the add negative to the uppers cotan^2x and a negatibe out of integral so u will have the result sinh-1(cotanx)
so the final answe is sinx+sinh-1(cotanx)+c and if u need help i have 4 ways for integrating cscx my mail is sherif.el-sayed@student.guc.edu.eg