A cool trigonometric integral

Here's one that looks scary due to quartic powers but made light work off.
My complex analysis lectures:
• Complex Analysis Lectures
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[@chronoxe]

no way bro didn't use Feynman's trick

[@pritamsarkar1429]

Bro started feeling generous and so he came up with thos integral 😂🙏🏻

[@MichaelDruggan]

You don't need residues at the end. You just need to know the derivative of the arctan function

[@MrWael1970]

Very nice approach. Thank you for your effort.

[@edmundwoolliams1240]

Very traditional method! When I saw the thumbnail I thought you were going to do some very clever complex analysis😊

[@maths_505]

Nah not all the time mate😂

[@Salvador964]

Excelente explicación, gracias por compartir.

[@7yamkr]

Finally some maths 505 integrals that i can solve😧👍..

[@holyshit922]

Looks easy Maybe double angle at the beginning (cos^2(x)+sin^2(x))^2 = cos^4(x)+sin^4(x) + 2cos^2xsin^2x 1 = cos^4(x)+sin^4(x) + 1/2*sin^2(2x) cos^4(x)+sin^4(x) = 1-1/2*sin^2(2x) 1+cos^4(x)+sin^4(x) = 2 - 1/2*sin^2(2x) 1/(1+cos^4(x)+sin^4(x)) = 1/(2 - 1/2*sin^2(2x)) 1/(1+cos^4(x)+sin^4(x)) =2/(3+1-sin^2(2x)) 1/(1+cos^4(x)+sin^4(x)) =2/(3+cos^2(2x)) 1/(1+cos^4(x)+sin^4(x)) =2/(cos^2(2x)(4+3tan^2(2x))) t = tan(2x) But with this substitution maybe indefinite integral first

[@zachbills8112]

I got to the 4-sin^2(x) the same way, then split that up in partia1 fractions in sin(x). I then used the tangent half angle substitution and contour integration to finish.

[@VishalDubey-pp7zc]

0 to inf 1+t²/2(t⁴+t²+1) 1+1/t²/2(t²+1/t²+1) 1+1/t²/2((t-1/t)²+3) 0 to inf dz/z²+3 1/√3Arc tan(z/√3) π/2√3 Video recommended randomly on yt but nice questions on channel btw will explore after finishing my exams

[@CM63_France]

Hi, "ok, cool" : 6:15 , "terribly sorry about that" : 1:00 , 3:13 , 4:38 .

[@renesperb]

Another way to start is to use that sinx^4+cosx ^4= 1/4 cos[4 x]+3/4 . Then , after changing to the variable y = 4 x one has an integral which can easily be calculated using the well known substituion y= tan(t/2) .This leads to dy = 2/(1+t^2) dt and cos y = (1-t^2)/(1+t^2) , and this leads to a standard integral .

[@fabreze1257]

How do you get that first identity or is that just one I need to know?

[@renesperb]

@@fabreze1257 One way to get it is to write cos x = 1/2*( e^(i x) + e ^(- i x)) , and sin x = 1/2 i *( e^(i x) - e ^(- i x)) . If you calculate 4 th powers and add up and simplify you get the desired identity.

[@xizar0rg]

Why does this need complex analysis (residues)? Isn't the last integral just clearly arctan?

[@user-ex7fq9dy5e]

Yeah I agree he's polly just trolling

[@spiderjerusalem4009]

More exposure to the audiences

[@maths_505]

I was just joking my friend 😂

[@giuseppemalaguti435]

La funzione integranda diventa (1/2)1/(1-(sin2x/2)^2)..poi uso 2x=t,e tg(t/2)=u...uso gli integrali razionali I=π√3/6

[@theelk801]

once you got to the 0 to pi you could have changed it to -pi to pi and applied the residue theorem

[@jejnsndn]

Try to share a hard gemoetry problems

[@Inge_M]

why can you change theta by pi/2-theta without touching anything else? btw good content, keep it up!

[@mcalkis5771]

Kamal is just trolling us with those thumbnail colours.

[@henrymarkson3758]

This is the first integral on this channel that can be solved using high school maths. (The phase shift from theta -> theta - pi/2 changes the limits of integration, right?)

[@maths_505]

Well those limits remain the same when you switch the upper and lower ones after the transformation.

[@henrymarkson3758]

Yes and no. More No than Yes. The substitution flips the limits of integration and and the value of the integral is (supposed to be) reversed. But in this PARTICULAR SITUATION ONLY, the result is unaffected, because the cosine function is even, and a further substitution theta-> minus theta restores the value of the integral.

[@lakshay3745]

​@@henrymarkson3758No he's right , limit are always unaffected when replaced x-> a+b-x where a and b are the limits

[@henrymarkson3758]

Fair enough, I forgot about the old King's Rule. We'll forgive our friend Kamal just this once:)

[@Jalina69]

I will try it myself first 🙄

[@shivamdahake452]

Woke up to an integral that my grade 12 ass can actually solve. Today is a good day fellas !

[@maths_505]

Some variation is important 😂

[@shivamdahake452]

​​@@maths_505 I am just waiting to get it over with JEE advanced next month, I want to learn a lot more about calculus. Do you have any recommendations ? I have a good grasp on high school calculus and wish to learn more.

[@abdulllllahhh]

would a weierstrass sub work here

[@maths_505]

Yes indeed

[@user-fg2mf1wc5k]

it feels good to see nice calculations without complex calculus math

[@highplayz]

More easier way is to multiply by sec⁴x in numerator and denominator then let tanx = u

[@aravindakannank.s.]

i did the same😂