Trig Integration (1 Viewer)

[atakach99]

Find the integral (primitive function) of

sin squared 3x


plz help

 

[tommykins]

Find the integral (primitive function) of

sin squared 3x


plz help

You should rote-learn that cos^2 ax = 1/2 [ x + cos2ax ] and sin^2 ax = 1/2 [ x - cos2ax ].

 

[atakach99]

mmmm.

i know that integral of sin squared x dx = 1/2x - 1/4 sin 2x

so do i jsut substitute x=3x ????

 

[benji6667]

I solved this one using substitution:

sin^2 3x

sub 3x with u

becomes:

sin^2 u

Differentiate u = 3x and u get du = 3dx

1/3 F sin^2 3x . 3dx

substituting u becomes

1/3 F sin^2 u .du

Now we intergrate:

1/3(1/2U - 1/4sin2u) +C

Substitue u with 3x

= 1/3(1/2*3x - 1/4sin2*3x) +C

= 1/3(3x/2 - 1/4sin6x) +C

= 3x/6 - 1/12sin6x + C

And take out 1/2 as common factor (so it looks like the answer in Maths in Focus book 2)

= 1/2(x - 1/6sin6x) +C

 

[atakach99]

yea thats soo long


i would like to use Charity's way but i didnt really understand how she got it..

 

[tommykins]

mmmm.

i know that integral of sin squared x dx = 1/2x - 1/4 sin 2x

so do i jsut substitute x=3x ????

For the +- cos part, it is double the angle in the original function given (ie. cos^2 6x makes it a +12x in the brackets)

 

[nick90]

lol If you're forgetting your trig identities then you are buckley's chance of remembering how to integrate by parts.

 

[conics2008]

just use u=3x

then du/dx=3 therefore dx=1/3 du

then find 1/3 S sin^2 (u)

1/3 { 1/2 u - 1/4 sin2u }

sub u back in

you get 1/2x-1/6sin6x

that was simple.