Another thing which i suppose can be done is substituting x=cos2t This leaves us with 1-cos2t and 1+cos2t in the denominator and in the numerator we get the derivative of cos 2t since dx = -2sin2t dt Now 1-cos2t = 2 sin^2 t and 1+cos2t = 2 cos^2 t and on simplifying the entire expression we'll simply get sec^4 t in the numerator that can now be easily solved by taking sec^2t out separately and writing the other power as (1+tan^2 t)
Way to stick with it! If instead of updating the bounds with each substitution, you back substituted instead, the integral is a little nicer. It's just -(x+2) sqrt(1-x^2) / 3(x+1)^2, that's the antiderivative of the original without thinking about bounds. Plugging 0 and 1/2 into there is not so bad (at least given how the final answer is, you still have to play with some sqrt(3) in places)
Ha! Thanks Doron! You’re right I was exhausted after this one. I recorded method 2 and it’s definitely quicker but plenty of interesting things in this one. 👍👍👍
Another thing which i suppose can be done is substituting x=cos2t
This leaves us with 1-cos2t and 1+cos2t in the denominator and in the numerator we get the derivative of cos 2t since dx = -2sin2t dt
Now 1-cos2t = 2 sin^2 t and 1+cos2t = 2 cos^2 t and on simplifying the entire expression we'll simply get sec^4 t in the numerator
that can now be easily solved by taking sec^2t out separately and writing the other power as (1+tan^2 t)
Nice solution! Seems pretty easy that way :)
x=sinθ,I=int(1/(1+sinθ)^2)..θ=0,π/6...
You did a great job! That was pretty unwieldy. It will be interesting to see how the other method compares.
thanks Mike! I just finished recording method 2 this morning and I can tell you its less work for me for sure :)
Way to stick with it!
If instead of updating the bounds with each substitution, you back substituted instead, the integral is a little nicer.
It's just -(x+2) sqrt(1-x^2) / 3(x+1)^2, that's the antiderivative of the original without thinking about bounds.
Plugging 0 and 1/2 into there is not so bad (at least given how the final answer is, you still have to play with some sqrt(3) in places)
Thanks David. Good idea! Just evaluating the bounds is probably the longest part so cleaning that up would save some trouble.
Excellent and clear video👏👏👏A lot of work... You should get some rest... 😀
Ha! Thanks Doron! You’re right I was exhausted after this one. I recorded method 2 and it’s definitely quicker but plenty of interesting things in this one. 👍👍👍
Fantastic
Thanks! :)
I=int[0,1/2](1/sqrt(1-x^2)(1+x)^2)dx
x=sin(t)
dt=dx/sqrt(1-x^2)
I=int[0,pi/6](1/(1+sin(t))^2)dt
t->pi/2-t
I=int[pi/3,pi/2](1/(1+cos(t))^2)dt
u=t/2
dt=2du
I=1/4•int[pi/6,pi/4](sec^4(u))du
b=tan(u)
db=sec^2(u)du
I=1/4•int[1/sqrt(3),1](1+b^2)db
I=(b/4+b^3/12)|[1/sqrt(3),1]
I=(18-5sqrt(3))/54
thanks Max
Is the other method x = (1-u)/(1+u) ? This is another overpowered secret weapon like the Weierstrass substitution and it works very nicely on this one
Hi Samir. No but it is a substitution and video is coming soon. How would that work with (1-u)(1+u)?
@@owl3mathIf you substitute x = (1-u)/(1+u), after finding dx and everything the integral simplifies to (1/4) * (1/3 to 1)∫(u^(1/2) + u^(-1/2))du