integration help (1 Viewer)

[want2beSMART]

this is integration by t-method

1. the integral of d@/cos@

2. the integral of dx/(1- sinx)

3. the integral of dx/(1+sinx)

4. the integral of cos@/(2 - cos@) d@

5. the integral of cosx/(sin^2@ + 1) dx

6. the integral of tanx/ (1 + cosx) dx

 

[KFunk]

1. ∫ dθ/cosθ let t = tanθ/2 ---> dθ = 2dt/(1+t²)
cosθ = (1-t²)/(1 + t²)

int. = ∫ (1+t²)/(1-t²). 2/(1+t²) dt
= ∫ 2/(1+t)(1-t) dt
= ∫ 1/(1-t) + 1/(1+t) dt
= -ln(1-t) + ln(1+t) +c
= ln[(1+t)/(1-t)] +c
= ln[(1+tanθ/2)/(1 - tanθ/2)] +c

 

[KFunk]

2. ∫ dx/(1-sinx) let t = tanx/2 ---> dx = 2dt/(1+t²)
sinx = 2t/(1 + t²)

int. becomes
∫ 1/[1 - 2t/(1 + t²)].2dt/(1+t²)
= ∫ 1/[(1 - 2t + t²)/(1+t²].2dt/(1+t²)
= ∫ 2dt/(1 - 2t + t²)
= ∫ 2dt/(t-1)²
= -2/(t-1) = -2/(tanx/2-1)

something worth noting is that t² -2t + 1 can be written as (1-t)² as well as (t-1)² and this would yield an alternative answer of 2/(tanx/2-1) (positive). I'm not sure what to do about this.

For question 3 do pretty much the same process as the above.

 

[KFunk]

Have a go at some of the others and remember you can constuct a right angled triangle with sides (1) and (t) and hypotenuse √(1 + t²) to work out values of sin(x/2) and cos(x/2) ----> hence you can work out cosx, sinx and tanx since:
cosx = cos²(x/2) - sin²(x/2)
sinx = 2sin(x/2)cos(x/2)

I'll work though no. 4 though since it's a bit trickier at the end... in a sec after I've fed myself.

 

[KFunk]

4. ∫ cosθ/(2 - cosθ ) dθ let t = tanθ/2 ---> dθ = 2dt/(1+t²)
cosθ = (1-t²)/(1 + t²)

Integral becomes
= ∫ [(1-t²)/(1 + t²)]/[2 - (1-t²)/(1 + t²)] x 2/(1+t²) dt

collect 2 - (1-t²)/(1 + t²) to make (1 +3t²)/(1 + t²) and cancel stuff out:

=∫ (2 - 2t²)/(1 +3t²)(1 + t²) dt

= ∫ 4/(3t²+1) - 2/(t² +1) dt

= ∫ 4/3.1/(1/3 + t²) - 2/(t²+1) dt

= 4[√3]/3tan^-1([√3].t) - 2tan^-1t

subbing in t= tan(θ/2)
= 4[√3]/3tan^-1([√3].tan(θ/2)) - 2tan^-1tan(θ/2)

= 4[√3]/3tan^-1([√3].tan(θ/2)) -2θ/2

= 4[√3]/3tan^-1([√3].tan(θ/2))-θ + C

 

[richz]

just use tan@/2 =t then draw the triangle and use double angle results..

 

[HayleeKate]

I havent done intetgration yet but feel left out.. so im going to tell you what i know:
tanhx ={ e^x - e^-x } / {e^x + e^ -x }
sinhx = { e^x - e^-x } / 2
coshx + {e^x + e^-x} / 2

coshx is the shape of the wire hanging between 2 telegraph poles.

Thankyou. I have been of no use to anyone looking for help on integration, or any help to anyone looking for help on any HSC maths topic, I know it. But I enjoyed this.

 

[Slidey]

Hyperbolic trigonmetry isn't in the HSC course.

But they hint at it sometimes with things like: "integrate e^x-1/e^x" ( or 2sinh(x), whose integral is 2cosh(x) )
Which is: e^x+1/e^x = 2cosh(x) as expected.

Something else to note is that sin(iz)=sinh(z)

Derivation:
z=cos(x)+isin(x)=e^(ix)
1/z=cos(x)-isin(x)=e^(-ix)
z+1/z=2cos(x)=e^(ix)+1/e^(ix)
cos(x)={e^(ix)+1/e^(ix)}/2
Et cetera. For hyperbolic trig, a=ib in cos(a), so:
cos(a)={e^(ia)+1/e^(ia)}/2
{e^(i*ib)+1/e^(i*ib)}/2
={1/e^(b)+e^(b)}/2
cosh(b)={e^(b)+1/e^(b)}/2

 

[HayleeKate]

I knew it wasnt in the syllabus, but I just find maths so thrilling I wanted to bring it up anyway.
I wasnt aware of the sin(iz)=sinh(z) property.. thats very cool, thanks Steele! [awesome]

 

[Captain pi]

6.

∫ ^tan(x) / _{1 + cos(x)} dx

Let t = tan(½x)
=> dt = ½sec²(x/2) dx
and, hence,
dx = 2cos²(x/2) dt

Now, considering a triangle described by t = tan(x/2) – that is, a triangle with adj. side = 1 where opp. side = t and hence hypotenuse² = 1 + t².

Thus,

cos²(x/2) = 1 / (1 + t²)

Thus, plugging into

∫ ^tan(x) / _{1 + cos(x)} dx

maps to

∫ ^{2t / _{(1 – t²)}} / _{1 + ^2t / (1 + t²)} × 2 / (1 + t²) dt

= ∫ ^{4t / _{(1 – t²)}} / _{1 + t² + 2t} dt

= ∫ ^{4t / _{(1 – t²)}} / _{1 + t² + 2t} dt

= ∫ ^4t / _{(1 – t²) (1 + t)²} dt

= Resolution into partial fractions.

trivial