√(1+sin(x)) is a tricky function since it's not differentiable at 3π/2 + 2kπ for k∈ℤ Accordingly, even with the correct antiderivative, calculating a definite integral may need to be done piecewise.
1+sinx=1+cos(x-π/2)=2cos²(x/2-π/4) [1+cosα=2cos²(α/2)] √(1+sinx)=√2cos(x/2-π/4)(if cos(x/2-π/4)≥0) So the integral of the function is equal to 2√2sin(x/2-π/4)+C
Yup, I spotted that at 2:55. For a channel that's supposed to teach correct practices in calculus, that's pretty bad. Did you see I credited your answer in one of my comments on a recent video?
Thank you. Unfortunately, there are errors that appear to be correct, but people do not notice them. For example, if someone writes √x²=x, it is considered correct, but if he writes √x²=-x, any person will see it as incorrect, knowing that the probability of being correct or incorrect in each of them is 50%.@@d-hat-vr2002
√(1+sin(x)) is a tricky function since it's not differentiable at 3π/2 + 2kπ for k∈ℤ
Accordingly, even with the correct antiderivative, calculating a definite integral may need to be done piecewise.
1+sinx=1+cos(x-π/2)=2cos²(x/2-π/4) [1+cosα=2cos²(α/2)]
√(1+sinx)=√2cos(x/2-π/4)(if cos(x/2-π/4)≥0)
So the integral of the function is equal to
2√2sin(x/2-π/4)+C
Solution:
∫√(1+sin(x))*dx =
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Substitution:
u = 1+sin(x) sin(x) = u-1 cos(x) = √(1-(u-1)²) = √(1-(u²-2u+1)) = √(2u-u²)
du = cos(x)*dx dx = 1/cos(x)*du
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= ∫√u*1/√(2u-u²)*du = ∫√[u/(2u-u²)]*du = ∫√[1/(2-u)]*du =
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Substitution: z = 2-u dz = -du du = -dz
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= -∫1/√z*dz = -∫z^(-1/2)*dz = -2*z^(1/2)+C = -2*√(2-u)+C
= -2*√{2-[1+sin(x)]}+C = -2*√{2-1-sin(x)}+C = -2*√[1-sin(x)]+C
Checking the result by deriving:
{-2*√[1-sin(x)]+C}’ = 2*1/2*1/√[1-sin(x)]*cos(x)
= cos(x)/√[1-sin(x)] = √[1-sin²(x)]/√[1-sin(x)]
= √{[1-sin²(x)]/[1-sin(x)]} = √{[1+sin(x)]*[1-sin(x)]/[1-sin(x)]}
= √(1+sin(x))
This result is only true in the interval sin(x/2)+cos(x/2)≥0 because √(a²)=a if a≥0.
Yup, I spotted that at 2:55. For a channel that's supposed to teach correct practices in calculus, that's pretty bad.
Did you see I credited your answer in one of my comments on a recent video?
Thank you. Unfortunately, there are errors that appear to be correct, but people do not notice them. For example, if someone writes √x²=x, it is considered correct, but if he writes √x²=-x, any person will see it as incorrect, knowing that the probability of being correct or incorrect in each of them is 50%.@@d-hat-vr2002
Exactly!