Integrate(sqrt (5-4x-x^2 ))

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Prime Newtons

3 ай бұрын

In this video, I showed how to compute an indefinite integral of the square-root of a quadratic

Пікірлер: 35

@holyshit922 3 ай бұрын

It is possible to calculate it by parts with clever choice of integration constant u = sqrt(5-4x-x^2) , dv = dx du = -(2+x)/sqrt(5-4x-x^2)dx , v = (x+2)

@PrimeNewtons 3 ай бұрын

You always have a great substitute. There's something I learned from you, I will be using soon. Thanks.

@hassanejturay2994 3 ай бұрын

@@PrimeNewtonshow can I reach you, I have something to discuss with you

@rithvikarun7112 3 ай бұрын

How would u integrate -(x+2)^2/sqrt(5-4x-x^2) dx Nvm got it as an answer with cos-1(x+2/3)

@holyshit922 3 ай бұрын

@@rithvikarun7112 I would rewrite it as -(x+2)^2=9-(5-4x-x^2) and I would have situatioon I = (x+2)sqrt(5-4x-x^2)+9Int(1/sqrt(5-4x-x^2),x) - I

@kushagrasharma5983 3 ай бұрын

Sir could you teach how to solve cubic equations? (Not by hit and trial method) 🙏🙏

@rithvikarun7112 3 ай бұрын

Yes please

@abhishankpaul 3 ай бұрын

Cardan's method is one way. That is taught in UG courses. There is a very nasty formula though for direct calculations, you won't get any formula over 4th order polynomial. Mathematics restricts you upto that only

@boguslawszostak1784 3 ай бұрын

@@abhishankpaulThe method is not as nasty as it seems at first glance, but the result is often nasty and rarely useful for substitution into other formulas. In the first step, you substitute x=y+z, multiply and choose z so that the coefficient next to y^2 is equal to 0. Then You move y^3 to one side and the rest to the other side, obtaining the equation to solve: y^3=py+q. The second step is to substitute y=u+v. But You don't substitute it into the equation, but use the formula for (u+v)^3, which can be written as (u+v)^3= 3uv*(u+v)+u^3+v^3, Now you insert y= u+v into this formula and obtain y^3=3uvy+(u^3+v^3). The third step is to choose u and v in such a way as to obtain from the general formula from step 2 the equation being solved, i.e., p=3uv, q=(u^3+v^3). Now we calculate v from the first equation v=p/(3u) and substitute it into the second one: q=(u^3+(p/3u)^3)=u^3+p^3/(27u^3). This is an equation with an unknown u^3, which can be denoted by t. q=t+p/(27*t). This is a quadratic equation, from which we only need one solution. The equation u^3=t has three solutions, each of which gives you one solution to your equation.

@user-wh2kn1bp9q 3 ай бұрын

Very good explanation. Thank you very much 🙂

@stigastondogg730 3 ай бұрын

Ace of Base 10 - “I saw the sine”

@bazboy24 3 ай бұрын

Amazing video

@surendrakverma555 3 ай бұрын

Thanks Sir

@antonionavarro1000 3 ай бұрын

Para que la expresión sqrt{5 - 4x -x^2} tenga sentido, se debe cumplir que 5 - 4x -x^2 >= 0 5 - 4x -x^2 >= 0 factorizando - (x+5) (x-1) >= 0 (x+5) (x-1) = 0) and (x-1 =-5 and x= 0 y por tanto cos(theta) >= 0. En consecuencia su valor absoluto no provoca ningún cambio, | cos(theta) | = cos(theta) No es que en el cambio de variable x-2 = 3•sin(x) se exija que sin(x) sea positivo, sino que, de hecho, ya lo es, atendiendo a su dominio.

@cliffordabrahamonyedikachi8175 3 ай бұрын

Thank you sir. Long answer.

@nullplan01 3 ай бұрын

Actually, I just did multiple substitutions. From Cardano's method, I know that it is possible to perform a linear substitution in a polynomial to lose the second highest term. A bit of playing around shows that in this case, the substitution x = t - 2 manages that, replacing the radicand with 9-t². Well, that is almost what I want, but not quite, so the second substitution I did was t = 3s. Because then the radicand becomes 9-9s², so I can bracket out the 9 and take its root separately. It joins the three from the fact that dt = 3 ds, and becomes a 9 in front of the integral. So now I have I = 9 ∫√(1-s²) ds And now finally I can do the trigonometric substitution I was aiming for all this time: Let s = sin u, ds = cos u du I = 9 ∫√(1-sin² u) cos u du = 9 ∫ cos² u du = 9 ∫ 1/2 (1 + cos 2u) du = 9/2 (u + 1/2 sin 2u) = 9/2 (u + sin u cos u) If s = sin u, cos u = √(1-s²) I = 9/2 (arcsin s + s √(1-s²)) = 9/2 (arcsin t/3 + t/3 √(1 - t²/9)) = 9/2 (arcsin (x+2)/3 + (x+2)/3 √(1 - (x+2)²/9)) + C

@biswambarpanda4468 3 ай бұрын

Long live sir..

@Aaryamank132 Ай бұрын

Thanks you sir 🙏🙏🙏. You are a great teacher. I am living in india .

@happyhippo4664 2 ай бұрын

I like how the parenthesis after 9/2 mysteriously appeared. The chalk board has autocorrect.😀

@vinayakmamtani 3 ай бұрын

We can just memorise the result for √a²-x²=x/2√a²-x² +a²/2sin-(x/a)

@Mephisto707 3 ай бұрын

Theta can’t be on the 1st quadrant only, because you will not hit every x in the domain of the function. Theta must be in the 1st AND 4th quarters. It just so happens cosine is positive on those 2 quadrants.

@PrimeNewtons 3 ай бұрын

Correct. Now I have to make a video explaining what you just said.

@abhishankpaul 3 ай бұрын

@@PrimeNewtons don't delete this video if you make a new one. Provide link of this video over there and new video's link here

@jumpman8282 3 ай бұрын

@@PrimeNewtons When you do, keep in mind that 9 = (±3)², which means that (𝑥 + 2)² ∕ 9 = (±(𝑥 + 2) ∕ 3)². Also, −sin 𝜃 = sin(−𝜃). Thereby, with sin 𝜃 = (𝑥 + 2) ∕ 3, we can let 𝜃 ∈ Q1 because the negative sign will take care of the angles in Q4.

@levysarah2954 3 ай бұрын

On t aime Sir Newton.continue a nous régaler avec tes vidéos

@hassanejturay2994 3 ай бұрын

❤

@okmotivated4786 3 ай бұрын

Brother it's my humble request that please solve this question If the last three 3 digits of x⁴ is (x-58)², then what is the last digit of x is?

@korayoduncu5883 3 ай бұрын

The only thing i did not understand was how that cos square tetha turnes into (1/2 + 1/2 cos2theta)

@mil9102 3 ай бұрын

Trig identity

@rogerkearns8094 3 ай бұрын

mil9102's right and there are proofs in textbooks. We just learn the formula and use it when we need to. The one for sin squared is half minus half cos 2 theta.

@kevinmadden1645 3 ай бұрын

Cos(2 theta) equals 2(cos squared theta)-1. Solve for (cos squared theta) and the result is readily obtained.

@user-hl7lt8ge7o 3 ай бұрын

(X+2)/3 is variety from -INF to +INF. Sin(Tetha) is variety from -1 to +1. Therefore I don’t understand how is possible to equivalent (x+2)/3=Sin(Tetha). Please discribe me it.