About NUR ARINA BASILAH BINTI KAMISAN Expertise i can answer derivatives and integral questions but not so expert in answering limits,vector and infinite series.
Experience i teached calculus for foundation students at university malaya and i also teached tutorial class in university for subject differential equation when i was doing my degree.currently i'm teaching algebra for foundation students.
Education/Credentials i have degree in mathematics with education and now i'm futhering my study in master of statistics at university malaya.
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!!
