Відео

1:57 Does anyone remember in which chapter the trick of multiplying numerator and denominator by sec²x is taught to simplify sin²x rationals? It's so unfamiliar!

sin dx/ dx = 1, similarly, sin (dx)²/(dx)²=1， therefore, sindx/sin(dx)² =dx÷(dx)²=1/dx=inf

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muy buen desarrollo aunque algo rebuscado ,,,,,,,

Very nice😮😮😮😮😮😮

6:45 The integration rule can only be used if we replace the variable t=x -1/2

Solution: ∫x²/(2*x²+1)*dx = ∫x²/[(√2*x)²+1]*dx = -------------- Substitution: u = √2*x x = u/√2 du = √2*dx dx = du/√2 -------------- = 1/(2*√2)*∫u²/[u²+1]*du = 1/(2*√2)*∫[(u²+1)-1]/(u²+1)*du = 1/(2*√2)*{∫(u²+1)/(u²+1)*du-∫1/(u²+1)*du} = 1/(2*√2)*{u-arctan(u)+C} = 1/(2*√2)*u-1/(2*√2)*arctan(u)+C/(2*√2) = 1/(2*√2)*√2*x-1/(2*√2)*arctan(√2*x)+D = 1/2*x-1/(2*√2)*arctan(√2*x)+D Checking the result by deriving: [1/2*x-1/(2*√2)*arctan(√2*x)+D]’ = 1/2-1/(2*√2)*1/[(√2*x)²+1]*√2 = 1/2-1/{2*[(√2*x)²+1]} = {[(√2*x)²+1]-1}/{2*[(√2*x)²+1]} = 2*x²/{2*[2*x²+1]} = x²/(2*x²+1) everything okay!

Solution: 2 ∫(3-4x)³*dx = -1 ------------------ Substitution: u = 3-4x du = -4*dx dx = -1/4*du upper limit = 3-4*2 = -5 lower limit = 3-4*(-1) = 7 ------------------ -5 7 7 = -1/4*∫u³*du = 1/4*∫u³*du = 1/16*[u^4] 7 -5 -5 = 1/16*[7^4-(-5)^4] = 111

(secx)^2/((secx)^2+(tgx)^2)=(secx)^2/(1+2(tgx)^2)..[t=tgx]..dt/(1+2t^2)=(1/2)dt/(t^2+1/2)=>(√2/2)arctg√2t=(1/√2)arctg(√2tgx)

Where does supposing x to be equal 3sin(theta) come from?

Multiply numerator and denominator by sec^2(x) and the rest will follow.

Solution: 1 ∫x*(1-x)^n*dx = 0 --------------------- Substitution: u = 1-x x = 1-u du = -dx dx = -du upper limit = 1-1 = 0 lower limit = 1-0 = 1 --------------------- 0 1 1 = -∫(1-u)*u^n*du = ∫[u^n-u^(n+1)]*du = [u^(n+1)/(n+1)-u^(n+2)/(n+2)] 1 0 0 = [1^(n+1)/(n+1)-1^(n+2)/(n+2) = 1/(n+1)-1/(n+2) = [1*(n+2)-1*(n+1)]/[(n+1)*(n+2)] = [n+2-n-1]/[(n+1)*(n+2)] = 1/[(n+1)*(n+2)]

√(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.

I tried to find an easier way, but your way is the easiest and most beautiful. May God bless you and grant you health and wellness.

not by partial fractions

1. Solution Integration by parts with u=x , dv = (1-x)^{n}dx 2. Solution Substitution u = 1-x

Integration by parts solves the problem but you avoid it with all costs x^4+1 = (1-x^4)+2x^4 and separate integral into two integrals then integral -\int{\frac{\sqrt{x^4-1}}{x^2}}dx calculate by parts with u = \sqrt{x^4-1} , dv = -\frac{1}{x^2}dx Answer is \frac{\sqrt{x^4-1}}{x}+C

（(x^4 ＋ 1)/(2x sqrt(x^4 - 1））+ c

2:10 Wrong. There is nothing in the question to suggest that x must be positive, so √(x²) = |x| , not x. Looking at the graph of the integrand, it is positive everywhere it's defined. So any interval where it's defined will have a positive definite integral. But if we try to use the expression boxed at 8:14 to calculate the area under the curve from (for example) -20 to -10 we get a (wrongly) negative answer. This video is a good example of what can go wrong when not thinking clearly about roots of powers! It seems the more intuitive method would be by substitution and working the +x and -x branches as separate integrals. Curiously, if he hadn't stopped at 8:14 but repeated the same mistake by re-writing the final answer as (√(x²-1)) / x he would accidentally found the correct antiderivative. Even so, the fact that the function is undefined on [-1,1] should lead to caution in computing definite integrals.

the background tone ! likee it 3:01

Another interesting problem is the generalisation : x^p(1-x)^q for natural p and q. Edit : Spoilers - (You can solve this entire class of integrals for a fixed p+q)

Solution: ∫√(1+sin(x))*dx = ------------- 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 ------------- = ∫√u*1/√(2u-u²)*du = ∫√[u/(2u-u²)]*du = ∫√[1/(2-u)]*du = ------------- Substitution: z = 2-u dz = -du du = -dz ------------- = -∫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.

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

Beta function

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My initial reaction to this video was not positive. Manipulating the expression to set up for the substitution u = x + 1/x at 3:10 requires a clever sequence of steps that a normal person wouldn't think to do. Thus this video has limited value as a teaching tool of generally useful methods of integration. at 0:21 few people would think to make the expression look more messy by factoring and cancelling x² and then at 1:30 few people would think to rewrite the numerator with +1 = +2 -1 and then at 2:02 few people would think to rewrite that 2 as 2⋅x⋅(1/x) That being said, this is not an easy integral using textbook methods most people are familiar with. The "correct" way to work this question is by partial fractions, by factoring the denominator as (x²+x+1)(x²-x+1). Alas, few people would recognize that the denominator could be factored thus. But an earlier commenter provided a technique that normal people would be more likely to notice: x⁴+x²+1 looks a lot like x⁴+2x²+1 which can be written as (x²+1)² So one needs to use only a single "trick", as the other commenter shows: x⁴+x²+1 = x⁴+2x²+1-x² = (x²+1)² - x² That last form is the well-known difference of two squares giving the desired factorization (x²+x+1) ⋅ (x²-x+1) Of course, fully working out the partial factor decomposition requires some algebra, a system of 4 linear equations to find the fractions' numerators, and a couple basic substitution integrals. But once the denominator is factored, it's just a matter of working carefully, not genius.

Very well done, good tutorial. For completeness, one should include calculations for the special cases of n=-1 and n=-2 since the initial question did not exclude these values which make the general answer undefined.

When you integrated 1/x with IBP, you left out the absolute value with the ln. Your answer is only true if x>0 and that is the issue I have with the solution.

Instead of suppose, you could have directly used the property f(x) = f(a-x)

that wwas smoth, and im proud i could solve it before seeing the video !

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❤ I = integral of (1/x - ln x)/e^x now we have to put denominator in squared form I = integral of { e^(x) (1/x) - ln x (e^x) } e^(2x) I = (ln x)/(e^x) + c

Solution: ∫1/(√x+∛x)*dx = ------------------ Substitution: u = x^(1/6) x = u^6 √x = x^(3/6) = u³ ∛x = x^(2/6) = u² du = 1/6*x^(-5/6)*dx dx = 6*x^(5/6)*du = 6*u^5*du ------------------ = 6*∫u^5/(u³+u²)*du = 6*∫u³/(u+1)*du = ------------------ Substitution: t = u+1 u = t-1 dt = du ------------------ = 6*∫(t-1)³/t*dt = 6*∫(t-1)*(t²-2t+1)/t*dt = 6*∫(t³-3t²+3t-1)/t*dt = 6*∫(t²-3t+3-1/t)*dt = 6*(t³/3-3t²/2+3t-ln|t|+C) = 2t³-9t²+18t-6*ln|t|+6C = 2*(u+1)³-9*(u+1)²+18*(u+1)-6*ln|u+1|+D = 2*(u³+3u²+3u+1)-9*(u²+2u+1)+18*(u+1)-6*ln|u+1|+D = 2u³+6u²+6u+2-9u²-18u-9+18u+18-6*ln|u+1|+D = 2u³-3u²+6u-6*ln|u+1|+D+11 = 2*√x-3*∛x+6*x^(1/6)-6*ln|x^(1/6)+1|+E

Integrali razionali...I=(1/2)ln{(x^2-x+1)/(x^2+x+1)}

t=x^(1/6)...=>I=2t^3-3t^2+6t-6ln(1+t)=2√x-3x^(1/3)+6x^(1/6)-6ln(1+x^(1/6))+c...dovrei ricontrollare...

(x²-1)/(x⁴+x²+1)=(x²-1)/(x⁴+2x²+1-x²)=(x²-1)/((x²+1)²-x²))=(x²-1)/((x²+x+1)(x²-x+1))=(1/2)[(2x-1)/(x²-x+1)-(2x+1)/(x²+x+1)] So the integral of the function is equal to (1/2)[ln|(x²-x+1)/(x²+x+1)|]+C

ngl, this was easy

This video is poor. First problem: From 0:19 to 1:47 No normal person would think to re-write the numerator like that. I have mentioned polynomial long division as the go-to first step for many integrals of this form in several previous videos. Why doesn't Mr. Calculus Booster show polynomial long division here, so that viewers can become familiar with the more general approach to this kind of question, rather than just magically re-writing the numerator with no explanation of how he thought to do that? The integral that begins to be written at 2:58 would be the form arrived at by polynomial long division. Second problem: At 6:18 he quotes a standard result from an integral table (here, #17 from the edition of Stewart that I have) But as in several previous videos he fails to include the step of checking that the integral, in doing the desired substitution, still matches the standard form. (here, he should do the step of substitution u=x-1/2 and then du=dx)

noice

dx/x is just dlnx, so the integral is equivalent to the integral of dlnx/lnx which is ln llnx l+C

A shorter way to solve this integral is when you get to the partial fractions section use the coverup method cause both x's have the same power.

Solution: ∫1/[x*ln(x)]*dx = ------------------ Substitution: u = ln|x| x = e^u du = 1/x*dx dx = x*du ------------------ = ∫1/(x*u)*x*du = ∫1/u*du = ln|u|+C = ln|ln|x||+C

solve the limit in factorization it' so good

I did this integral in a few seconds in my head because i used this trick i found out that x^3 is equal to (x*sqrt(x))^2 and sqrt(x)*dx = (2/3)*d(x*sqrt(x)).

Solution: lim[(16-x²)/(x²+x-12)] = x➞-4 ----------------- Secondary calculation: x²+x-12 = 0 |p-q-formula ⟹ x1/2 = -1/2±√(1/4+12) = -1/2±√(49/4) = -1/2±7/2 ⟹ x1 = -1/2+7/2 = 3 and x2 = -1/2-7/2 = -4 ⟹ x²+x-12 = (x-x1)*(x-x2) = (x+4)*(x-3) ------------------ = lim{(4-x)*(4+x)/[(x+4)*(x-3)]} x➞-4 = lim[(4-x)/(x-3)] = 8/(-7) = -8/7 x➞-4

I did it by using partial fraction decompostion and my answer was slightly different from yours but only in the constant which was -(1/5)ln(2) cause i had t+3 in the denominator instead of 2t+6 = 2(t+3).

If the limit gives you 0 over 0 then one can apply de L'Hôpital's rule which results into -8/7.

in rational expressions, when taking the limit, if its factorable, factor it and cancel the common factors, and then you use the substitution. Im preparing and getting ready to study Basic Calculus next semester which is next month on february, the 2nd semester.

Solution: lim{[√(x²+7)-4]/(x-3)} = [using the rule of l’Hospital] = x➞3 lim{[1/2*1/√(x²+7)*2x]/1} = x➞3 lim[x/√(x²+7)] = 3/4 x➞3