Trigonometric Integrals --- ∫sin^n(x)cos^m(x)dx via Pythagorean or Half-Angle Identities

[@uchiyamataichi3202]

1/4sin4x also has 1/2 so I think we need to multiply 1/2 by the last term and it'll be "1/32sin4x". Anyway thank you for giving me a nice method of integral.

[@DrTrefor]

Thanks for that!

[@calcubite9298]

The identities above his head at 5:14, I am told are called "reduction identities", because you are reducing the power of the exponent, and half-angle identities are a separate thing. The professor may want to make a video describing the differences (and similarities) between the reduction identities, the multiangle (e.g., double angle) identities, and the fractional (e.g., half-angle identities). Writing this comment sent me down a Wikipedia rabbit hole, starting with en.wikipedia.org/wiki/List_of_trigonometric_identities#Half-angle_formulae

[@user-uc7tw1kg8i]

thank you for this series fr

[@naiko1744]

I'd say the big take away here is that all integrals of the form sin(x)]^n * [cos(x)]^m can be solved in the same two ways. Furthermore, whenever we are despaired, we can apply trig identities.

[@gokulseshadri3455]

Trefor Sir!! You are the best!! Nowadays I am gaining my lost interest in Math again

[@holyshit922]

Maybe Pythagorean identity then Chebyshev polynomial, then system of linear equations

[@ElifArslan-l9g]

Thank you

[@AlBoulley]

sin²(x) cos²(x) -- I suggest Pythagorean sub immediately after FOIL of “half‑angle” subs, producing ¼ sin²(2x), which becomes ⅛ (1 - cos4x) and THAT integrates with less effort… resulting in ⅛ (x - ¼ sin4x) + c You strike me as thorough and the kind of person that (almost always) catches their own mistakes. Thus, I'm surprised you skipped simplifying your answer to that integral because doing so would have caught what became an error.

[@crimson4066]

How does (1/4)((1/2)(1+cos(4x)) turn into (x/8)-(1/16)sin(4x) and not (x/8)-(1/8)sin(4x)?

[@ChrisFranklin.2260]

final answer of second problem is 1/4 x - 1/32 x * sin*(4x) + C not 1/16

[@Chrisymcmb]

Indeed I was a bit confused since I was following along on paper

[@teoeconomu6601]

I think it is 1/8 x - 1/32 sin(4x) + C

[@thehiddenworld751]

i think we can euler forumla to help linearsing the sin and cos

[@physicslover1950]

Please make a video on trigonometric integration of powers of tan and secant when power of tan is even and power of secant is odd. But power of tan should not be 2 and power of secant should not be one

[@carultch]

Example: integral tan(x)^4*sec(x)^3 dx Split the sec(x)^3 into sec(x)^2 and sec(x): integral tan(x)^4*sec(x)^2 * sec(x) dx Use tan(x)^2 = sec(x)^2 - 1. Thus tan(x)^4 = (sec(x)^2 - 1)^2. integral (sec(x)^2 - 1)^2 * sec(x)^2 * sec(x) dx Expand: integral sec(x)^7 - 2*sec(x)^5 + sec(x)^3 dx Now we have an integral that is only in terms of secant. These are all parts integrals with looper stops. I'll skip ahead to the results: integral sec(x)^7 dx = 1/768*(198*sin(x) + 85*sin(3*x) + 15*sin(5*x))*sec(x)^6 + 5/16*atanh(sin(x)) integral sec(x)^5 dx = 1/32*(11*sin(x) + 3*sin(3 x))*sec(x)^4 + 3/8*atanh(sin(x)) integral sec(x)^3 dx = 1/2*sec(x)*tan(x) + 1/2*atanh(sin(x)) Combine the above to get the solution: 1/768*[78*sin(x) - 47*sin(3*x) + 3*sin(5*x)]*sec^6(x) + 1/16*arctanh(sin(x)) + C

[@viktorashistan7207]

haha he said long and hard

[@soundbytes6564]

🥺