integral of sqrt(1-4x^2) ,Trig Substitution, calculus 2 tutorial

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Integral of sqrt(1-4x^2), Trig Substitution, Stewart Calc 7th ET ed
Trig Substitution, calculus 2 tutorial.
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#blackpenredpen #math #calculus #apcalculus

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I love how you go through each step so that it's all so easy to understand. Thank you, and keep up the good work!

Dank_SomeOne

Genius! Brilliant! Exactly what I was looking for, and much more comprehensive and understandable than most other videos/forum threads on this integral!

TimeTraveler-hkxo

THX from Brazil !!!! love ur guides ! thx

pauloborges

√(1-4x²): x belongs to [-1/2, 1/2]，so when x = 1/2 sinθ than -1 <= sin θ <= 1, assume -1/2 π <= θ <= 1/2 π, then cos θ >= 0, so √(1-sin² θ) = cos θ;

exlife

Hi sir I am watching your video in India. ❤❤ Thanks sir for help solve to query questions

Sachinvermafzd

Hola hermano . Soy mexicano y veo tus videos desde ya hace un tiempo gracias a ti pase mi materia de cálculos integral. Gracias mil gracias

josueobettcoelloperez

thanks for also putting the section # and problem #! it makes the problem easy to find, this of course assuming everyone is using the same textbook! thank you!

MissStuckOnBandaids

thank you !! This video is 6 years old but still very relevant in 2021.

nebiyathamza

Thanks. I love to watch your tutorials. Great job.

rifaathanver

Found an another way to solve this question.
Let x = sinycosy
On differentiating both sides we get:
dx = cos(2y)dy
Now back to the question:
∫sqrt(1 - 4x^2)dx
= ∫sqrt(1 - 4sin^2ycos^2y)•cos(2y)dy
= ∫cos^2(2y)dy [After simplifying]
= (1/2)∫1 + cos(4y)dy
= (1/2)y + (1/8)(sin4y)
= (1/2)y + (1/4)sin2y•cos2y ...(1)
Now, we know that x = sinycosy
Therefore, 2x = sin(2y) ...(2)
After constructing a right-angled ∆, we get:
cos(2y) = sqrt(1 - 4x^2) ...(3)
And from equation (2), we get:
y = (1/2)arcsin(2x) ...(4)
Putting the values of sin(2y), cos(2y) and y from equations (2), (3) and (4) respectively to equation (1), we get:
(1/2)y + (1/4)sin2y•cos2y
= (1/4)arcsin(2x) + (1/2)x•sqrt(1-4x^2)
Hence,
∫sqrt(1 - 4x^2)dx
= (1/4)arcsin(2x) + (1/2)x•sqrt(1-4x^2) + C
Thank you..

日本人じゃありません

Thanks. Just started this. Reading through.

josephshaff

Thank you so much! I'm doing physics in college and never really got integration (apart from the basics) and you've really helped!

ellerose

Thank you so much for these videos! They are invaluable and you explain very well.

briannabyrd

You could also substitute u=2x (, because 4x^2=(2x)^2), so that you get 1/(2sqrt(1-u^2))du, which integrated is 1/2arcsinu+C=1/2arcsin2x+C

georgiyotov