Expert: NUR ARINA BASILAH BINTI KAMISAN - 7/23/2008

Question
1. cosecx cotx/1 + cosecx (dx) 2. (y-2)^2 /y4 (dy)
3.tan^2 (3x-2) dx

Answer
dear betty kazunge,

1.for your first question,u must first know what is cosecx and cotx.
cosecx=1/sinx and cotx=cosx/sinx
from above,substitute it into the original equation and you will got:

(cosecx cotx)/(1+cosecx)dx
=(1/sinx*cosx/sinx)/(sinx/sinx*1/sinx)dx
=(cosx/sin^2x)/(sinx+1/sinx)dx
=(cosx/sin^2x)*(sinx/sinx+1)dx
=cosx/(sinx(sinx+1))dx
by letting u=sinx and du=cosx dx,we will have
=du/u(u+1)
and by doing part by part procedure we will turn the equation into
=[1/u-1/u+1] du
integrate the above equation and we will get
=ln(u)-ln(u+1)+c
and as we know u=sinx,so the equation will be
=ln(sinx)-ln(sinx+1)+c or
=ln(sinx/sinx+1) + c

2.for your 2nd question.first you must expand (y-2)^2 to y^2-4y+4 and substitute to the original equation and you will get

(y-2)^2/y^4dy
=(y^2-4y+4)/y^4 dy
=(y^2/y^4) - (4y/y^4) + (4/y^4) dy
=(1/y^2) - (4/y^3) + (4/y^4) dy
=(y^-2) - (4y^-3) + (4y^-4) dy
integrate the equation above
=(-y^-1)+(2y^-2)-(4/3y^-3)+c
=(-1/y)+(2/y^2)-(4/3y^3)+c

3.tan^2(3x-2)dx
let 3x-2=u and du=3dx
substitute into the original equation and we'll have
=tan^2(u)/3 du
and from identity of trigonometry tan^2(x)=sec^2(x) - 1
=1/3(sec^2(u) - 1)du
integrate equation above (integration of sec^2x=tanx)
=1/3[tan u - u] + c
and u=3x-2 and we will get the answer
=1/3[tan (3x-2) - (3x-2)] + c

-->try to do it on a piece of paper and understand it..good luck ya!!