วีดีโอ

Thanks for an other video master.

How did you factor a pi/2 on 2:33? I think you made too many mistakes

Also you may solve woth Feynman taknic

Wow, just now I realised you could use the d(cosx) notation for IBP 🤯🤯

Great video

Rewrite as \int{\frac{e^{x}}{1+\cos{x}} dx} + \int{\frac{e^{x}\sin{x}}{1+\cos{x}} dx} Integrate by parts \int{\frac{e^{x}\sin{x}}{1+\cos{x}} dx} with u = \frac{\sin{x}}{1+\cos{x}} , dv = e^{x}dx and combine integrals

I see integration by parts here with u = ln(x)

τ

an interesting integral

π ln (2) / 2

Using kings rule would solve it a lot quicker 🙌

Wow I would never have thought of it this way! I split it into integrals between -π~0 and 0~π and it turns into (∫[-π, 0] (x + |x|)sinx dx + ∫[-π, 0] (x + |x|)sinx dx), and now since x in the first integral is always negative, you can write |x| as -x, and the same logic allows you to write |x| as x in the second integral. So you end up with ∫[-π, 0] (x - x)sinx dx + ∫[-π, 0] (x + x)sinx dx which is ∫[-π, 0] 0 · sinx dx + ∫[-π, 0] 2x · sinx dx = 0 + 2 · ∫[-π, 0] sin dx = 0 + 2π = 2π

Your writing is so wonderful.❤❤❤❤❤

Solution: ∫1/(5^x+1)*dx = ∫(5^x+1-5^x)/(5^x+1)*dx = ∫(5^x+1)/(5^x+1)*dx-∫5^x/(5^x+1)*dx = ∫dx-∫5^x/(5^x+1)*dx = x-∫e^[x*ln(5)]/{e^[x*ln(5)]+1}*dx = x-1/ln(5)*∫e^[x*ln(5)]/{e^[x*ln(5)]+1}*ln(5)*dx --------------------- Substitution: u = e^[x*ln(5)]+1 du = ln(5)*e^[x*ln(5)] --------------------- = x-1/ln(5)*∫1/u*du = x-1/ln(5)*ln|u|+C = x-1/ln(5)*ln|e^[x*ln(5)]+1|+C = x-1/ln(5)*ln|5^x+1|+C Checking the result by deriving: [x-1/ln(5)*ln|5^x+1|+C]’ = 1-1/ln(5)*1/(e^[x*ln(5)]+1)*e^[x*ln(5)]*ln(5) = 1-5^x/(5^x+1) = (5^x+1-5^x)/(5^x+1) = 1/(5^x+1) everything okay!

I would split the fractions. First lets work on ploynomial on denominator X^4=-1 X^2=+-i X^2=i => x1=√2/2(1+i) x2=-√2/2(1+i) X^2=-i => x3=√2/2(1-i) X4=-√2/2(i-i). Now x1*x3=1/2(1+1)=1 and x1+x3=√2 X2x4=1 and x2+x4=-√2 So x^4+1=(x^2+√2x+1)(x^2-√2x+1) (Ax+B)/(X^2+√2x+1)+(Cx+D)/(X^2-√2x+1)=X^2/(x^4+1) (Ax+B)(X^2-√2x+1)+(Cx+D)(X^2+√2x+1)=X^2 X=√2/2(1+i) => (D+C*√2/2(i+1))*(i+1+i+1)=+i D(2+2i)+2√2C*i=i D=0 C=1/2√2 Similarly then A=0 B=-1/2√2 => X^2/(x^4+1)=√2/4(1/x^2-√2x+1)-1/(x^2+√2x+1))=√2/4(1/((x-√2/2)^2+(√2/2)^2)-1/((x+√2/2)^2+(√2/2)^2)) Now all that is left is using the identity Arctan'(x/a)=a/(x^2+a^2) I missed an x on nunerator so we need to first make the top similar to derivative of denominator first by afdind and subtracting a number which will resilt in logarithm antiderivative and then the arcatngent formula

a good part of mathematic

Just take divide numerator and denominator by 5^x. You get integral of dx/5^x(1+1/5^x) Then you set u=1/5^x, du=ln(1/5)/5^x. dx Substitute into the equation 1/ln(1/5) × integral of du/1+u. And the answer is ln(1+1/5^x)/ln(1/5)

an interesting integral

14‏/11‏/2024

an interesting integral

Buteful 🎉🎉🎉

an interesting integral

Thanks, if the integral had no upper and lower limits, how would it be solved, Professor?

Thanks for an other video master

Thanks

an interesting integral

an interesting integral

You amazing me with those internal half angles

an interesting integral

an interesting integral

Thanks for an other video master

an interesting integral

an interesting integral

I=int((xsec^2(x)+tan(x))/(1+xtan(x))^2)dx t=1+xtan(x) dt=(xsec^2(x)+tan(x))dx I=int(t^-2)dt I=-1/t+C I=-cos(x)/(cos(x)+xsin(x))+C

I=int((x^-2-cos(x))/(1/x+sin(x))^2)dx t=sin(x)+1/x dt=(cos(x)-x^-2)dx I=int(-t^-2)dt I=1/t+C I=x/(1+xsin(x))+C

an interesting integral

an interesting integral

I got e^x(1+cos x)/sin x + c

an interesting integral

an interesting integral

an interesting integral

an interesting integral

Integration by parts

What is the name of the technique you used?

Buteful

an interesting integral

Thanks for an other video master

an interesting integral

an interesting integral

an interesting integral