Integral of the Day 8.31.24 | Trig Integral—Calculus 2 | Math with Professor V

Here's your latest Integral of the Day! Hope you enjoy! Did you solve it differently? Comment down below!

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Professor V

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[tonychow]

Hi Professor V, thank you for your new video. I did something different, I multiplied denominator and numerator by tanx and let u=secx. I got the integral of u^2/u^2-1. My final result was not the same as yours, but I have diffrentiated my result and got the original integral. I tried Daniel's method too and indeed it was faster. Cheers Professor V and Daniel.

[ryanrasco]

do have a video doing an integral the long way, sort of like doing a derivative using limit of the difference quotient, but do the integral using Riemann sum notation?

P.s. These have been SO much help thank you!

[danielweitsman]

I reduced the integrand to csc(x)sec^2(x) and then did integration by parts, where U=csc(x), dV=sec^2(x). Then it went pretty quickly.

[shvnshavan]

Hi professor v can you assist me in solving this problem, integrate √(1+4cos²(2y)) dy from 0 to π/2
Thank you

[physicsmajor]

I was like uh..

sec³(x) / tan(x)
= sec²(x) × sec(x) × tan(x) / tan²(x)
= [sec²(x) / (sec²(x) – 1)] × sec(x) tan(x)

So if we let u = sec(x), then du = sec(x) tan(x) dx, and becomes

int [ u² / (u² - 1)] du
= int du + int 1/(u² - 1) du
= u + 1/2 ln| (u-1)/(u+1) | + C
= sec(x) + 1/2 ln| (sec(x) - 1)/(sec(x) + 1) | + C

Please correct me if you found something wrong. Thanks.