Question

How to do this integral? trigonometric sustitution.

Answer 1

\[\int\limits_{a}^{b}(4-y)\sqrt{4-y ^{2}}dy\]

Answer 2

This one is a little bit trickier than the other one, the other one was pretty easy in comparison I believe, I'm lacking the first steps only I guess...

Answer 3

Use the substitution:

\[\sqrt{a^2-y^2}=>y=asin(\theta)\]

Answer 4

But do I first multiply 4-y and the root? I need a little push on the very first steps, I can take it from then on.

Answer 5

Take the integral:
integral (4-y) sqrt(4-y^2) dy
For the integrand (4-y) sqrt(4-y^2), substitute y = 2 sin(u) and dy = 2 cos(u) du. Then sqrt(4-y^2) = sqrt(4-4 sin^2(u)) = 2 cos(u) and u = sin^(-1)(y/2):
= 2 integral 2 (4-2 sin(u)) cos^2(u) du
Factor out constants:
= 4 integral (4-2 sin(u)) cos^2(u) du
Write cos^2(u) as 1-sin^2(u):
= 4 integral (4-2 sin(u)) (1-sin^2(u)) du
Expanding the integrand (4-2 sin(u)) (1-sin^2(u)) gives 2 sin^3(u)-4 sin^2(u)-2 sin(u)+4:
= 4 integral (2 sin^3(u)-4 sin^2(u)-2 sin(u)+4) du
Integrate the sum term by term and factor out constants:
= 8 integral sin^3(u) du-16 integral sin^2(u) du-8 integral sin(u) du+16 integral 1 du
Use the reduction formula, integral sin^m(u) du = -(cos(u) sin^(m-1)(u))/m + (m-1)/m integral sin^(-2+m)(u) du, where m = 3:
= -8/3 sin^2(u) cos(u)-8/3 integral sin(u) du-16 integral sin^2(u) du+16 integral 1 du
The integral of sin(u) is -cos(u):
= (8 cos(u))/3-8/3 sin^2(u) cos(u)-16 integral sin^2(u) du+16 integral 1 du
Write sin^2(u) as 1/2-1/2 cos(2 u):
= (8 cos(u))/3-8/3 sin^2(u) cos(u)-16 integral (1/2-1/2 cos(2 u)) du+16 integral 1 du
Integrate the sum term by term and factor out constants:
= (8 cos(u))/3-8/3 sin^2(u) cos(u)+8 integral cos(2 u) du+8 integral 1 du
For the integrand cos(2 u), substitute s = 2 u and ds = 2 du:
= (8 cos(u))/3-8/3 sin^2(u) cos(u)+4 integral cos(s) ds+8 integral 1 du
The integral of cos(s) is sin(s):
= (8 cos(u))/3+4 sin(s)-8/3 sin^2(u) cos(u)+8 integral 1 du
The integral of 1 is u:
= 4 sin(s)+8 u+(8 cos(u))/3-8/3 sin^2(u) cos(u)+constant
Substitute back for s = 2 u:
= 8 u+4 sin(2 u)+(8 cos(u))/3-8/3 sin^2(u) cos(u)+constant
Apply the double angle formula sin(2 u) = 2 sin(u) cos(u):
= 8 u+(8 cos(u))/3-8/3 sin^2(u) cos(u)+8 sin(u) cos(u)+constant
Express cos(u) in terms of sin(u) using cos^2(u) = 1-sin^2(u):
= 8 u+8 sqrt(1-sin^2(u)) sin(u)+(8 cos(u))/3-8/3 sin^2(u) cos(u)+constant
Substitute back for u = sin^(-1)(y/2):
= -1/3 sqrt(4-y^2) y^2+2 sqrt(4-y^2) y+(4 sqrt(4-y^2))/3+8 sin^(-1)(y/2)+constant
Which is equal to:
8 sin^(-1)(y/2)-1/3 sqrt(4-y^2) (y^2-6 y-4)+constant

Answer 6

hmm do this \[\int(4-y)\sqrt{4-y^2}dy=\int 4 \sqrt{4-y^2}dy-\int y\sqrt{4-y^2}dy\]

Answer 7

the second one is a straight forward substitution

Answer 8

the first you gonna do trig sub like the dude did above

Answer 9

Okay, let me try real fast, thanks!

Answer 10

chaise already did the work lol

Answer 11

i have no will to check the work, but the start is good. so should be good