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# P3 Help Please? Couple Questions watch

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1. 1. ∫ xsec ²x

2. Solve the differential equation: dy/dx = (x^4-1)/(x ²y ²) given that y=2 when x=1
2. (Original post by Sang)
1. ∫ xsec ²x

2. Solve the differential equation: dy/dx = (x^4-1)/(x ²y ²) given that y=2 when x=1
1) ∫ xsec²x

v=x
dv/dx = 1
du/dx=sec²x
u=tanx

=> xtanx - ∫ tanx
=> xtanx - ln(secx)

NB - The ln(secx) bit is from member when I would normally use the formula sheet so you might want to check what the integral of tanx actually is!!!
3. (Original post by Leekey)

NB - The ln(secx) bit is from member when I would normally use the formula sheet so you might want to check what the integral of tanx actually is!!!
yeah it is....because INT tanx = -ln(cosx) = ln(cosx)^-1 = ln(secx)
4. (Original post by Sang)
1. ∫ xsec ²x

2. Solve the differential equation: dy/dx = (x^4-1)/(x ²y ²) given that y=2 when x=1
question 2:

y^2 dy = (x^4-1)/x^2 dx = x^2 - x^-2 dx

so 1/3 y^3 = 1/3 x^3 + x^ -1 + C

8/3 = 1/3 + 1 + C

C = 4/3

so y^3 = x^3 + 3x^-1 + 4
5. (Original post by Sang)
2. Solve the differential equation: dy/dx = (x^4-1)/(x ²y ²) given that y=2 when x=1
∫ y² dy = ∫ (x^4-1)/(x²) dx

You can take it from there...
6. thanks people

one question i dont understand is,

∫ xsec^x

u=x
du/dx=1
dv/dx=sec^2x
so v= lnsecxtanx

so i did, xlnsecxtanx - ∫ lnsecxtanx

then what? i was on the right track then wasn't i?
7. (Original post by Sang)
dv/dx=sec^2x
so v= lnsecxtanx
v = tanx
8. (Original post by Sang)
thanks people

one question i dont understand is,

∫ xsec^x

u=x
du/dx=1
dv/dx=sec^2x
so v= lnsecxtanx

so i did, xlnsecxtanx - ∫ lnsecxtanx

then what? i was on the right track then wasn't i?
umm if dv/dx is sec^2 x.... then v is tan x.

so your integral should turn out to be x tanx - integral of tan x.
9. ∫ xsec²(x) dx

f(x)=x ; g'(x)=sec²x=(tanx)'
f'(x)=1 ; g(x)=tanx

So, ∫ xsec²(x) = x*tanx - ∫1*tanx dx

=xtanx - ∫sinx/cosx dx

=xtanx + ∫(-sinx)/(cosx) dx

=xtanx + ∫(cosx)'/cosx dx

=xtanx + ln¦cosx¦ + k
10. ahh excellent,

my final question is this (more integration)

a) Show that d/dx [e^3x (cos2x+ksin2x)] = e^3x[(3k-2)sin2x + (2k+3)cos2x]

where k = constant

b) Letting k=2/3 find ∫ e^3x cos2x dx

c) Find ∫ e^3x sin2x dx

then im sorted, thanks!
11. a) Show that d/dx [e^3x (cos2x+ksin2x)] = e^3x[(3k-2)sin2x + (2k+3)cos2x]
differentiating using the product rule gives:

= -2e^(3x).sin(2x) + 3e^(3x).cos(2x) + 3ke^(3x).sin(2x) + 2ke^(3x).cos(2x)

= e^(3x).[sin(2x).(3k-2) cos(2x).(3+2k)]

b) Letting k=2/3 find ∫ e^3x cos2x dx
well when k = 2/3, you can substitute into the answer to (a) and then reverse the process in (a) to get the answer to the integral. Therefore substituting 2/3 gives the integral of e^(3x).cos2x and the answer is:

e^(3x).cos(2x) + (2/3).e^(3x).sin(2x)

c) Find ∫ e^3x sin2x dx
now substitute k = -2/3 to remove cos(2x) term from the answer to (a). Then using the full method to (a) reversed, the answer is:

e^(3x).sin(2x) - (2/3)e^(3x).sin(2x)

Hope this helps.

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Updated: June 1, 2004
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