Integral of sin^4x

This calculus video tutorial explains how to find the integral of sin^4x using the power reducing formulas and pythagorean identities found in trigonometry.

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[jimcar]

What ever your name is - You are the most informative U tube teacher i have listened to. GREAT JOB.

[rubasham]

i've never failed to find a video of yours on a subject that i'm having trouble with, thank you sooo much!

[AydinovaSabina]

Your videos are very useful. Greetings from Azerbaijan

[evelynluna]

You did an amazing job at teaching it thank you!

[joerivera]

Awesome you made that very simple. Thank you

[sulav.]

Thank uuuu, you're the bessst ❤️‍🔥❤️‍🔥❤️‍🔥

[piusokolo]

Please do you have a video on Jordan normal form

[oxygen]

Can i solve it by using intigration by part ?

[LukasKamin]

is the formula of sum squared, namely (a+b)^2=a^2+2ab+b^2 NOT taught in western schools that's not the first time I see this well-known formula from basic algebra neglected by this guy, it looks so weird to me. BTW I wonder does these formulas (of sum/difference squared as well as differnce of squares and the same with cubic powers) have some name in western mathematical tradition? in ex-USSR tradition we call them formulas of abbreviated/shortened multiplication (translated literally), for each of this formulas has one side that may be presented as the multiplication of the alike or same expressions in braces

[iijvhfn]

I used integration by part then u- substitution, but I got completely different answer

[spirosgrivas]

Nice work! Thank you very much! Interesting you call yourself The Organic Chemistry Tutor but you have so much Math and Physics in here. And what about Physical and Inorganic Chemistry? lol

[upeecb]

There is no other method instead of it

[who_we_are______]

Dude what's your name? Also will you ever do a face reveal?

[theowillis]

This is L from death note. for stem students/Ppl who do stem for fun/genuises

[TheLukeLsd]

I tried the hard way 😅

ʃsin²(x)sin²(x)dx
Let
u=sin²(x)
dv=sin²(x)dx

—>
du= 2sin(x)cos(x)dx

v=ʃsin²(x)dx
v=ʃsin(x)sin(x)dx

v=ʃrds

r=sin(x)
dr=cos(x)dx
ds=sin(x)dx
s=-cos(x)

v=rs-ʃsdr = =
-sin(x)cos(x)+ʃcos²(x) =
-sin(x)cos(x)+ʃ(1-sin²(x))dx=
sin(x)cos(x)+ʃdx - ʃsin²(x)dx = ʃsin²(x)dx
—> 2ʃsin²(x)dx=sin(x)cos(x)+ x —>
ʃsin²(x) =(sin(x)cos(x) + x )/2


—>ʃsin⁴(x)dx = ʃsin²(x)sin²(x)dx = ʃudv = uv-ʃvdu = u(rs-ʃsdr)-ʃ(rs-ʃsdr)du =
sin²(x)(rs-ʃsdr) - ʃ(rs-ʃsdr)2sin(x)cos(x)dx =

-
=
-

Let cos²(x)=1-sin²(x) & ʃsin²(x) =(sin(x)cos(x) + x )/2

-
=
-
=
+ x )/2] - + x )/2]2sin(x)cos(x)dx
=
-
=
-
=
3/2×sin²(x)[-sin(x)cos(x)+x]
=
3/2×{sin²(x)[-sin(x)cos(x)+x]
=
3/2×{sin²(x)[-sin(x)cos(x)+x] + 2×ʃsin²(x)cos²(x)dx + 2ʃxsin(x)cos(x)dx}

Let dn=sin²(x)cos(x)dx & m=cos(x)

ʃsin²(x)cos²(x)dx = ʃmdn = mn-ʃndm =
= sin³(x)cos(x) + ʃsin⁴(x)dx

Then
3/2×{sin²(x)[-sin(x)cos(x)+x] + 2×ʃsin²(x)cos²(x)dx + 2ʃxsin(x)cos(x)dx}
=
3/2×{sin²(x)[-sin(x)cos(x)+x] + 2×(sin³(x)cos(x) + ʃsin⁴(x)dx) + 2ʃxsin(x)cos(x)dx}

Let p=x & dq=sin(x)cos(x)dx
ʃxsin(x)cos(x)dx = ʃpdq = pq-ʃqdp =
xsin²(x)/2-ʃsin²(x)/2×1dx =
½xsin²(x)-½ʃsin²(x)dx =
½xsin²(x)-½(sin(x)cos(x) + x )/2 =
½xsin²(x)-¼(sin(x)cos(x) + x )

Then
3/2×{sin²(x)[-sin(x)cos(x)+x] + 2×(sin³(x)cos(x) + ʃsin⁴(x)dx) + 2ʃxsin(x)cos(x)dx}
=
3/2×{sin²(x)[-sin(x)cos(x)+x] + 2×(sin³(x)cos(x) + ʃsin⁴(x)dx) + 2[½xsin²(x)-¼(sin(x)cos(x) + x )]}
=
3/2×{sin²(x)[-sin(x)cos(x)+x] + 2×(sin³(x)cos(x) + ʃsin⁴(x)dx) + xsin²(x)-½(sin(x)cos(x) + x )}
=
3/2×{sin²(x)[-sin(x)cos(x)+x] + 2sin³(x)cos(x) + 2ʃsin⁴(x)dx + xsin²(x)-sin(x)cos(x)/2 - x/2 }
=
3/2×{-sin³(x)cos(x)+xsin²(x) + 2sin³(x)cos(x) + 2ʃsin⁴(x)dx + xsin²(x)-sin(x)cos(x)/2 - x/2 }
=
3/2×{2xsin²(x) + sin³(x)cos(x) + 2ʃsin⁴(x)dx -sin(x)cos(x)/2 - x/2 }
=
3xsin²(x) + 3sin³(x)cos(x)/2 + 3ʃsin⁴(x)dx -3sin(x)cos(x)/4 - 3x/4
=ʃsin⁴(x)dx —>

3xsin²(x) + 3sin³(x)cos(x)/2 -3sin(x)cos(x)/4 - 3x/4
= -2ʃsin⁴(x)dx —>

2ʃsin⁴(x)dx= -3xsin²(x) - 3sin³(x)cos(x)/2 +3sin(x)cos(x)/4 + 3x/4
—>
ʃsin⁴(x)dx =3x/8 + 3sin(x)cos(x)/8 - 3sin³(x)cos(x)/4 - 3xsin²(x)/2 + C.
—>
ʃsin⁴(x)dx =3x/8 + 3sin(x)cos(x)/8 - 6sin³(x)cos(x)/8 - 12xsin²(x)/8 + C.
—>
ʃsin⁴(x)dx =3x(1-4sin²(x))/8 + 3sin(x)cos(x)(1-2sin²(x)/8 + C.

[aguilarkurtlaurenceralfg.]

Im so confusee. Can someone explain it to me why sin^2 (x) = 1/2 (1-cos 2 (x)) because i thought sin^2 (x) = 1-cos^2 (x). Hehe. Thank you. Im just too dumb for this.