:::Need help wit a few questions::: (1 Viewer)

[mazza_728]

Hey guys,

Im stuck on some questions, wondering if you could offer some help:

1 Int sin@/(2+sin@) d@

2 similarly the Int cos@/(2-cos@) d@

3 Int sinxcos2x dx

4 Int sin²2xcos²2x dx

Thanks xoxo

 

[Xayma]

Hmm There will most probably be a really quick easy way for 4 but you could:
standard sin² x etc
=1/4 int (1-cos 4x)(1+ cos 4x) dx
=1/4 int (1-cos² 4x) dx
=1/8 int (2-(1+cos 8x)) dx
=1/8 int (1-cos 8x) dx
=1/8 (x-1/8 sin 8x) +C
=x/8-1/64 sin 8x +C

 

[jm1234567890]

for 1 and 2, use t formula

for 3 expand the cos2x in terms of cos.

 

[mazza_728]

i used the t formula and got this god for saken fraction to integrate and i thought perhaps i could do it by partial fractions but it just got even more messy!

 

[jm1234567890]

for 4

((cos2x)^2)((sin2x)^2)
= (1/2(2cos2xsin2x))^2
= (1/2(sin(4x)))^2

which is alot easier.

 

[jm1234567890]

Originally posted by mazza_728
i used the t formula and got this god for saken fraction to integrate and i thought perhaps i could do it by partial fractions but it just got even more messy!

yeah, partial fractions is a pain, but sometimes you gota use it.


btw, how do you type powers easily?

 

[Xayma]

Originally posted by jm1234567890
which is alot easier.

Originally posted by Xayma
Hmm There will most probably be a really quick easy way for 4 but you could:

There is always an easier way then my answers

 

[mazza_728]

to type powers u just write < sup > 2 (or watever the power is) < / sup > with no spaces though. its really handy!!

The partial fractions chapter in fitz isnt that bad i thought but these partial fractions are impossible! e.g. Int 2t.dt/((t²+t+1)(1+t²))
or
Int (2-2t² . dt)/((1+t²)(1+3t²))

????

 

[ToO LaZy ^*]

test...<sup>2
EDIT: yay, it works

 

[CM_Tutor]

Originally posted by mazza_728
The partial fractions chapter in fitz isnt that bad i thought but these partial fractions are impossible! e.g. Int 2t.dt/((t²+t+1)(1+t²))

Both t² + t + 1 and t² + 1 are irreducibile quadratics, so to do the partial fractions we need to find constants A, B, C and D such that:

2t / [(t² + t + 1)(t² + 1)] = [(At + B) / (t² + t + 1)] + [(Ct + D) / (t² + 1)]

Eliminating fractions, we have: 2t = (At + B)(t² + 1) + (Ct + D)(t² + t + 1) _____ (*)

Taking t = 0 shows that B + D = 0 _____ (1)

Equating coefficients of t³ show that A + C = 0 _____ (2)

Equating coefficients of t² shows that 0 = B + D + C _____ (3)

Putting (1) into (3) gives C = 0, which can be put into (2) to give A = 0

So, (*) has become: 2t = B(t² + 1) + D(t² + t + 1)
Equating coefficients of t shows that D = 2, and using (1) we get B = -2.

So, 2t / [(t² + t + 1)(t² + 1)] = [-2 / (t² + t + 1)] + [2 / (t² + 1)]

The other one would be similar.

 

[mazza_728]

yuck i hate those kind of partial fractions. thankyou for helping me though .. much appreciated!

I think i may have to do these thru partial fractions as well but their just not working:

Int (x+2)/(x²-1)

and

Int x³/(x²+2x+1) i know i have to divide first which i did and ended up with Int (x-2) + (3x+2)/(x²+2x+1) how can i simplify this further? or do i need to use partial fractions?

and

Int x/(x²+2)²

Similarly Int dx/(x²+x)

Int 6/(9-x²)

Int sqt (a²-x²)/x²

????

Last one (for the moment ) i promise sorry guys! none of them seem to be happening for me! um
Int sin²xcos²x dx ive tried it both ways i know i.e. let cos²x=(1-sin²x) and also (sinxcosx)²=1/2sin2x but both ways dont give me the right answer the 2nd way i got close but still i checked all my working and it still didnt yeah happen.. please help!


Thanks again xoxo

 

[grimreaper]

Originally posted by mazza_728
Last one (for the moment ) i promise sorry guys! none of them seem to be happening for me! um
Int sin²xcos²x dx ive tried it both ways i know i.e. let cos²x=(1-sin²x) and also (sinxcosx)²=1/2sin2x but both ways dont give me the right answer the 2nd way i got close but still i checked all my working and it still didnt yeah happen.. please help!

how about using sinxcosx=1/2sin2x (I havent done 4u integration yet but I think thats what youre supposed to do there)

 

[mazza_728]

Originally posted by grimreaper
how about using sinxcosx=1/2sin2x (I havent done 4u integration yet but I think thats what youre supposed to do there)

i tried that :

MYSELF
Int sin2xcos2x dx ive tried it both ways i know i.e. let cos2x=(1-sin2x) and also (sinxcosx)²=1/2sin2x but both ways dont give me the right answer the 2nd way i got close but still i checked all my working and it still didnt yeah happen.. please help!

 

[George W. Bush]

(sin2xcos2x)^2 = (1/2 sin4x) ^2 = 1/4 sin^2 4x.

1/4 (sin^2 4x) = 1/4 (1/2 - cos8x/2)
= 1/8 (1 - cos8x)

integrating.. 1/8 (x - sin8x/8) + c

 

[CM_Tutor]

Originally posted by mazza_728
Int (x+2)/(x²-1)

Partial Fractions, (x + 2) / (x² - 1) = [A / (x + 1)] + [B / (x - 1)

Int x³/(x²+2x+1) i know i have to divide first which i did and ended up with Int (x-2) + (3x+2)/(x²+2x+1) how can i simplify this further? or do i need to use partial fractions?

(3x + 2) / (x² + 2x + 1) = (3x + 2) / (x + 1)² = [A / (x + 1)] + [B / (x + 1)²]

Int x/(x²+2)² dx

x / (x² + 2)² = [(Ax + B) / (x² + 2)] + [(Cx + D) / (x² + 2)²]

Similarly Int dx/(x²+x)

1 / (x² + x) = (A / x) + [B / (x + 1)]

Int 6/(9-x²)

6 / (9 - x²) = [A / (3 - x)] + [B / (3 + x)]

Int sqt (a²-x²)/x²

Is the x² inside the square root, or not?

Int sin²xcos²x dx ive tried it both ways i know i.e. let cos²x=(1-sin²x) and also (sinxcosx)²=1/2sin2x but both ways dont give me the right answer the 2nd way i got close but still i checked all my working and it still didnt yeah happen.. please help!

I agree with the posts above about using the sin2x result:
sin²x * cos²x = (sin x * cos x)² = [(1 / 2) * sin2x]² = (1 / 4) * sin²2x = (1/ 4) * (1 - cos4x) / 2 = (1 - cos 4x) / 8

 

[mazza_728]

oooo dont i love partial fraction
well thankyou very much guys.
manda xoxo