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.
Expert: NUR ARINA BASILAH BINTI KAMISAN - 8/13/2008
Question I'm trying to find general anti derivative of the following function
dy/dx= -9sin^5(3x)cos^2(3x)+12cos^4(3x)sin^3(3x)
i started by trying to anti derive in two parts
i couldn't get the first half but for the second half i got
dy/dx= 12cos^4(3x)sin^3(3x)
y= 12/9cos^3(3x)-12/15cos^5(3x)+c
please help"
Answer dear calum,
i'm not sure whether your second integration(anti derivative) is correct or not since i got a different answer from yours.could you please check back your answer?
when you have this kind of question we separate it first so you don't get confuse.let do the first half first ok.
let say dy/dx=-9sin^5(3x)cos^2(3x)
now change the equation into this equation:
dy/dx=-9[sin(3x)sin^4(3x)cos^2(3x)]
dy/dx=-9[sin(3x)(sin^2(3x))^2cos^2(3x)]
dy=3[-3sin(3x)(sin^2(3x))^2cos^2(3x)dx]
and from the trigonometry identity we know that sin^2(3x)=1-cos^2(3x)
substitute it into the equation above and we will have:
dy=3[-3sin(3x)(1-cos^2(3x))^2cos^2(3x)dx]
let u=cos(3x) then du=-3sin(3x)dx
substitute into the equation above and you will get:
dy=3[(1-u^2)^2*u^2]du
dy=3[(1-2u^2+u^4)*u^2]du
dy=3[u^2-2u^4+u^6]du
integrate with respect to u and we will get:
y=3[u^3/3-2u^5/5+u^7/7]
y=u^3-6u^5/5+3u^7/7]
substitute back u=cos(3x)
y=cos^3(3x)-6cos^5(3x)/5+3cos^7(3x)/7
and by using the same method as above,we will integrate the second half equation.
dy/dx=12[cos^4(3x)sin^3(3x)]
dy=12[(cos^2(3x))^2(sin^2(3x))sin(3x)]dx
dy=12[cos^2(3x)^2(1-cos^2(3x))sin(3x)]dx
let u=cos3x and du=-3sin(3x)dx
dy=12/-3[(u^2)^2(1-u^2)du]
dy=-4[u^4*(1-u^2)]du
dy=-4[u^4-u^6]du
integrate with respect to u and we will get:
y=-4[u^5/5-u^7/7]
y=-4u^5/5+4u^7/7 or
y=-4cos^5(3x)/5+4cos^7(3x)/7
combining both equation and we will have:
y=cos^3(3x)-6cos^5(3x)/5+3cos^7(3x)/7-4cos^5(3x)/5+4cos^7(3x)/7+c
and simplify it we will have:
y=cos^3(3x)-2cos^5(3x)+cos^7(3x)+c
try to write it in mathematical form and you will find it much more easy to understand.further information on this topic you can check this website: http://calc101.com/trig_powers.html
