2nd elegant method to evaluate the integral using differentiation under the integration sign

[@slavinojunepri7648]

I was very impressed by this solution the moment I read it as a comment in another video with the same integral. Thanks for this video post.

[@cipherunity]

Indeed, it is better than the 1st solution and you are welcome

[@AnityaBibhab]

Hey there 🤗 thanks for the video 😊. You got a new subscriber ❤.

[@Vedanshi-l4r]

❤

[@cipherunity]

Thanks to you. You solution is better than mine which I presented in the 1st video.

[@user-lu6yg3vk9z]

Have an idea for problem no method thinking about converting the cos(x)=cos(2*x/2)=cos^2(x/2)-sin^2(x/2) from there
((5)^1/2-(cos^2(x/2)-sin^2(x/2)))
((5)^1/2-(1-sin^2(x/2)-sin^2(x/2))
((5)^1/2-(1-2sin^2(x/2)))
sec^6(x/2)/(sec^6(x/2)((5)^1/2-1+2sin^2(x/2))^3

[@cipherunity]

What next

[@user-lu6yg3vk9z]

sec^4(x/2) sec^2(x/2)/((sec^2(x/2))^3(5^1/2-1+2sin^2(x/2))^3)
(tan^2(x/2)+1)^2 sec^2(x/2)/(sec^2(x/2)(5^1/2-1+2sin^2(x/2))^3
(tan^2(x/2)+1)^2 sec^2(x/2)/(5^1/2sec^2(x/2)-sec^2(x/2)+2tan^2(x/2))

[@holyshit922]

Use substitution
cos(x) = (1 - sin(x))t
cos^2(x) = (1 - sin(x))^2t^2
(1 - sin^2(x)) = (1 - sin(x))^2t^2
(1 - sin(x))(1 + sin(x)) = (1 - sin(x))^2t^2
1 + sin(x) = (1 - sin(x))t^2
1 + sin(x) = t^2 - sin(x)t^2
sin(x)+sin(x)t^2 = t^2 - 1
sin(x)(t^2 + 1) = t^2 - 1
sin(x) = (t^2 - 1)/(t^2 + 1)
cos(x) = (1 - (t^2 - 1)/(t^2 + 1))t
cos(x) = 2t/(t^2 + 1)
cos(x)dx = (2t(t^2+1)-2t(t^2-1))/(t^2+1)^2dt
cos(x)dx = 4t/(t^2+1)^2dt
2t/(t^2 + 1)dx = 4t/(t^2+1)^2dt
dx = 2/(t^2+1)dt
sqrt(5)-cos(x) = sqrt(5) - 2t/(t^2+1)
sqrt(5)-cos(x) = (sqrt(5)t^2 - 2t + sqrt(5))/(t^2+1)
But to use substitution we must divide interval of integration into two pieces
\int_{1}^{\infty}\frac{(t^2+1)^3}{(\sqrt{5}t^2 - 2t + \sqrt{5})^3}\cdot\frac{2}{(t^2+1)}dt
+\int_{-\infty}^{-1}\frac{(t^2+1)^3}{(\sqrt{5}t^2 - 2t + \sqrt{5})^3}\cdot\frac{2}{(t^2+1)}dt
2(\int_{-\infty}^{-1}{\frac{(t^2+1)^2}{(\sqrt{5}t^2 - 2t + \sqrt{5})^3}}dt + \int_{1}^{\infty}{\frac{(t^2+1)^2}{(\sqrt{5}t^2 - 2t + \sqrt{5})^3}}dt)
then use substitution
t - 1/sqrt{5} = \frac{2}{\sqrt{5}}u

[@user-lu6yg3vk9z]

How did u come up with substitution? cos(x)=(1-sin(x))t

[@holyshit922]

@@cipherunity
Did you differentiate
sin(x) = (t^2 - 1)/(t^2 + 1)
You have two ways
1. Quotient rule
2. Chain rule with algebraic manipulation
sin(x) = (t^2 - 1)/(t^2 + 1) = 1 - 2/(t^2+1)
I wrote it in the comment above
Because of existence x = pi/2 in interval of integration you should divide interval of integration into two subintervals
If you include that you will get right answer

[@holyshit922]

@@user-lu6yg3vk9z For simplicity assume that interval of integration is [0,pi/2]
We know that sqrt(1-w^2)=(1-w)t
rationalizes integrand
(we have rational parametrization of the curve y^2=1-x^2 this way)
If we plug in w = sin(x)
sqrt(1-sin^2(x))=(1-sin(x))t
Here I needed interval [0,pi/2]
because of square root
cos(x) = =(1-sin(x))t

[@cipherunity]

@@holyshit922 Every thing is clear except one thing. Explain your limits for 1 to infinity and -infinity to -1. What happen between -1 and 1

[@holyshit922]

@@cipherunity Problem is that in original interval there is point x = pi/2
which causes problems so we must divide interval into subintervals
cos(x)=(1-sin(x))t
so t = cos(x)/(1-sin(x))
and what happen for x=pi/2 which is inside of interval of integration