Integral of 1/(1+x^2)^(3/2) (substitution)

✍🏼 integralsforyo... - Integral of 1/(1+x^2)^(3/2) - How to integrate it step by step using integration by substitution!
✅ 𝐃𝐞𝐫𝐢𝐯𝐚𝐭𝐢𝐯𝐞 𝐭𝐨 𝐜𝐡𝐞𝐜𝐤 𝐭𝐡𝐞 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧
Derivative of x/sqrt(1+x^2) = • Derivative of x/sqrt(1...
🔧 𝐔𝐬𝐞𝐟𝐮𝐥 𝐯𝐢𝐝𝐞𝐨s
sin(arctan(x)) = • sin(arctan(x)) & cos(a...
🔍 𝐀𝐫𝐞 𝐲𝐨𝐮 𝐥𝐨𝐨𝐤𝐢𝐧𝐠 𝐟𝐨𝐫 𝐚 𝐩𝐚𝐫𝐭𝐢𝐜𝐮𝐥𝐚𝐫 𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐥? 𝐅𝐢𝐧𝐝 𝐢𝐭 𝐰𝐢𝐭𝐡 𝐭𝐡𝐞 𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐥 𝐬𝐞𝐚𝐫𝐜𝐡𝐞𝐫:
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[@IntegralsForYou]

🔍 𝐀𝐫𝐞 𝐲𝐨𝐮 𝐥𝐨𝐨𝐤𝐢𝐧𝐠 𝐟𝐨𝐫 𝐚 𝐩𝐚𝐫𝐭𝐢𝐜𝐮𝐥𝐚𝐫 𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐥? 𝐅𝐢𝐧𝐝 𝐢𝐭 𝐰𝐢𝐭𝐡 𝐭𝐡𝐞 𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐥 𝐬𝐞𝐚𝐫𝐜𝐡𝐞𝐫: ► Integral searcher 👉integralsforyou.com/integral-searcher 🎓 𝐇𝐚𝐯𝐞 𝐲𝐨𝐮 𝐣𝐮𝐬𝐭 𝐥𝐞𝐚𝐫𝐧𝐞𝐝 𝐚𝐧 𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐭𝐢𝐨𝐧 𝐦𝐞𝐭𝐡𝐨𝐝? 𝐅𝐢𝐧𝐝 𝐞𝐚𝐬𝐲, 𝐦𝐞𝐝𝐢𝐮𝐦 𝐚𝐧𝐝 𝐡𝐢𝐠𝐡 𝐥𝐞𝐯𝐞𝐥 𝐞𝐱𝐚𝐦𝐩𝐥𝐞𝐬 𝐡𝐞𝐫𝐞: ► Integration by parts 👉integralsforyou.com/integration-methods/integration-by-parts ► Integration by substitution 👉integralsforyou.com/integration-methods/integration-by-substitution ► Integration by trig substitution 👉integralsforyou.com/integration-methods/integration-by-trig-substitution ► Integration by Weierstrass substitution 👉integralsforyou.com/integration-methods/integration-by-weierstrass-substitution ► Integration by partial fraction decomposition 👉integralsforyou.com/integration-methods/integration-by-partial-fraction-decomposition 👋 𝐅𝐨𝐥𝐥𝐨𝐰 @𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐥𝐬𝐟𝐨𝐫𝐲𝐨𝐮 𝐟𝐨𝐫 𝐚 𝐝𝐚𝐢𝐥𝐲 𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐥! 😉 📸 instagram.com/integralsforyou/

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Needed this for proving field due to wire using coulombs law

[@IntegralsForYou]

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[@as_if6224]

sin(arctan(x)) is equal x/sqrt(x^2+1) because if we interpret x as the ratio between opposite and adjacent of an angle theta then we can create a right triangle with side lengths opposite=x, adjacent=1 and hypotenuse=sqrt(x^2+1) with arctan(x)=theta then sin(theta)=opposite/hypotenuse -> sin(theta)=x/sqrt(x^2+1)

[@IntegralsForYou]

Hi! Thank you for your comment! I did a video about it here kzbin.info/www/bejne/d2TIept4obGrZrs 💪

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[@IntegralsForYou]

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[@beogot]

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[@IntegralsForYou]

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[@mariabennardo8402]

Wonderful! I need to know how solve integral of 1/(1+×^2)^3, please could you help me?

[@IntegralsForYou]

Hi! Let's see how to solve it! Integral of 1/(1+x^2)^3 dx = Substitution: x = tan(u) ==> 1+x^2 = 1+tan^2(u) = 1/cos^2(u) dx = 1/cos^2(u) du = Integral of 1/(1/cos^2(u))^3 1/cos^2(u) du = = Integral of 1/(1/cos^6(u)) 1/cos^2(u) du = = Integral of cos^6(u)/cos^2(u) du = = Integral of cos^4(u) du = = integralsforyou.com/integrals?v=M_z1Jg-15R8 = = (1/32)( 12u + 8sin(2u) + sin(4u) ) = = (1/32)( 12u + 8*2sin(u)cos(u) + 2sin(2u)cos(2u) ) = = (1/32)( 12u + 16sin(u)cos(u) + 4sin(u)cos(u)cos(2u) ) = = (1/8)[ 3u + 4sin(u)cos(u) + sin(u)cos(u)cos(2u) ] = = (1/8)[ 3u + 4sin(u)cos(u) + sin(u)cos(u)( cos^2(u) - sin^2(u) ) ] = x = tan(u) ==> u = arctan(x) = (1/8)[ 3arctan(x) + 4sin(arctan(x))cos(arctan(x)) + sin(arctan(x))cos(arctan(x))( cos^2(arctan(x)) - sin^2(arctan(x)) ) ] = sin(arctan(x)) & cos(arctan(x)) ==> kzbin.info/www/bejne/d2TIept4obGrZrs = (1/8)[ 3arctan(x) + 4(x/sqrt(1+x^2))(1/sqrt(1+x^2)) + (x/sqrt(1+x^2))(1/sqrt(1+x^2))( (1/sqrt(1+x^2))^2 - (x/sqrt(1+x^2))^2 ) ] = = (1/8)[ 3arctan(x) + 4x/(1+x^2) + (x/(1+x^2))( 1/(1+x^2) - x^2/(1+x^2) ) ] = = (1/8)[ 3arctan(x) + 4x/(1+x^2) + (x/(1+x^2))((1-x^2)/(1+x^2)) ] = = (1/8)[ 3arctan(x) + 4x/(1+x^2) + (x-x^3)/(1+x^2)^2 ] = = (1/8)[ 3arctan(x) + 4x(1+x^2)/(1+x^2)^2 + (x-x^3)/(1+x^2)^2 ] = = (1/8)[ 3arctan(x) + (4x+4x^3)/(1+x^2)^2 + (x-x^3)/(1+x^2)^2 ] = = (1/8)[ 3arctan(x) + (4x+4x^3+x-x^3)/(1+x^2)^2 ] = = (1/8)[ 3arctan(x) + (3x^3+5x)/(1+x^2)^2 ] + C Hope it helped! ;-D

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[@IntegralsForYou]

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[@joejackson9738]

thank you so much it was my homework

[@IntegralsForYou]

My pleasure! ❤️❤️

[@jonpritzker9709]

HELP! Integrate xdx / sqrt [ ax^6 + bx^4 +cx^2 + d] I'm trying to do a double integral. Solved the inside with your help (this video), but it yielded this ^ beast, and I have no idea what to do. Furthermore, I have details about the constants. This came from trying to find the electric flux across the face of a dodecahedron if there's a proton at its center. The answer is q/12epsilon naught, because flux over whole dodecahedron is q/epsilon naught. But that's the easy way. This is the hard way... I think... This integral is proportional to the flux across part of the pentagonal face. Multiply it by 10 and also some other constants (dimensions of dodecahedron) and it SHOULD yield q/12epsilon. I would be SO happy to divulge info about constants upon request

[@Ramir0118]

¡Thanks for sharing!

[@IntegralsForYou]

My pleasure! 😊

[@kalles8789]

You can also use u-Substitution. Take 1/(1+x²)^3/2 and write it as 1/(1+x²) * 1/(1+x²)^1/2. then substitute u = 1/(1+x²). It is not difficult to transfer the whole Integral into the u-world. When it is in the u-world, substitute: s = (1-u)^1/2. After that you will have a very nice and easy polynome in s. Without any transformation into Cosine-function. Then substitute back. And simplify the expression of course. That is also not hard. This trick works with all integrals of 1/(1+x²)^(2n+1)/2 very nice.

[@IntegralsForYou]

Thank you very much for your comment! I put your solution in my TODO list and one day it will become a video! Thanks!

[@omer872]

@@IntegralsForYou you did it ?

[@omer872]

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[@IntegralsForYou]

@@omer872 Not yet! 🙃

[@omer872]

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[@itz_meaman8070]

Can't we take x² common and form x³ outside.. And then substitute (1/x² +1) = t?

[@IntegralsForYou]

Hi! I'm afraid we cant... well, you can, but it does not make the integral easier than before...

[@oCalippoTePiace]

TY❤!

[@IntegralsForYou]

My pleasure! 😊

[@gabriele21598]

What would be the solution if there was another number instead of the one? Thanks

[@IntegralsForYou]

Ciao Gabriele! You have to do the same substitution but this time it will be x/sqrt(a) = tan(u). Let's say we have "a" instead of the one: Integral of 1/(a+x^2)^(3/2) dx = = Integral of 1/(a(1+(x^2)/a)^(3/2) dx = = Integral of 1/[a^(3/2) (1+(x/sqrt(a))^2 )^(3/2) ] dx = = 1/a^(3/2) Integral of 1/[(1+(x/sqrt(a))^2 )^(3/2) ] dx = (*) Substitution: x/sqrt(a) = tan(u) ====> x = sqrt(a)*tan(u) dx = sqrt(a)*(1/cos^2(u))du (*) = 1/a^(3/2) Integral of 1/[(1+tan^2(u) )^(3/2) ] sqrt(a)*(1/cos^2(u))du = = sqrt(a)/a^(3/2) Integral of 1/[(1+tan^2(u) )^(3/2) ] (1/cos^2(u))du = = (1/a) Integral of 1/(1/cos^2(u))^(3/2) (1/cos^2(u))du = = (1/a) Integral of cos(u) du = = (1/a) sin(u) = = (1/a) sin(arctan(x/sqrt(a))) = = (1/a) (x/sqrt(a))/sqrt(1+(x/sqrt(a))^2) = = (1/a) x/sqrt(a+ x^2) + C By the way, I have to write it :D : Since there are two ones in this integral, I thought you were talking about the denominator. If not, it is a lot easier: Integral of a/(1+x^2)^(3/2) dx = = a Integral of 1/(1+x^2)^(3/2) dx = = ( watch the video again :D ) = = a*x/sqrt(1+x^2) + C Enjoy it!

[@aditisharma6728]

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[@IntegralsForYou]

You're welcome! Enjoy the channel! 😊😊💪

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you're legend

[@IntegralsForYou]

hehe thank you! 💪

[@aayushiahlawat3208]

[1/(1-x²) ³dx solution of this problem please

[@IntegralsForYou]

Hi! You have to do partial fraction decomposition for 1/(1-x^2)^3: 1/(1-x^2)^3 = 1/((1-x)(1+x))^3 = 1/((1-x)^3(1+x)^3) = A/(1-x) + B/(1-x)^2 + C/(1-x)^3 + D/(1+x) + E/(1+x)^2 + F/(1+x)^3 =...

[@phamthelam2005]

why sin(arctan(x))=x/(√1+x²)

[@IntegralsForYou]

Hi! I explained here kzbin.info/www/bejne/d2TIept4obGrZrs 😉

[@vincentm2325]

Hey ! Great video nut why did your X = tan ? Is that a formula that i don't know ? ^^. Thx

[@IntegralsForYou]

Hi Denis! Here we are doing the trigonometric substitution method: -For integrals that have sqrt(1-x^2) we can do x=cos(u) [dx=-sin(u)du] or x=sin(u) [dx=cos(u)du], so sqrt(1-cos^2(u))=sqrt(sin^2(u))=sin(u) or sqrt(1-sin^2(u))=sqrt(cos^2(u))=cos(u) Example: kzbin.info/www/bejne/iHmrcoqpbp15r7M Example: kzbin.info/www/bejne/b6bTmoqFoLyooKs - For integrals that have sqrt(1+x^2) we can do x= tan(u) [dx=1/cos^2(u) du], so sqrt(1+tan^2(u))=sqrt(1/cos^2(u)) = 1/cos(u) Example: kzbin.info/www/bejne/eqGVm56Gjbpsntk Example: kzbin.info/www/bejne/bKTWe5aKhreghLc -For integrals that have sqrt(x^2-1) we can do x=1/cos(u) [dx=sin(u)/cos^2(u)] so sqrt(1/cos^2(u) - 1)=sqrt(sin^2(u)/cos^2(u))=sqrt(tan^2(u))=tan(u)=sin(u)/cos(u) I think I have not an example yet, sorry...

[@firdausrizan8878]

Why we can't just use chain rule on this question?

[@IntegralsForYou]

Hi! How would you use it?

[@firdausrizan8878]

@@IntegralsForYou no sir, I'm just asking cause I'm really bad at integration lol

[@IntegralsForYou]

@Firdaus Rizan I asked because the chain rule is a derivation rule... 😉

[@jyotitiwari2138]

Integration of tan inverse X /(1+x)^2

[@IntegralsForYou]

Hi Jyoti Kumari! Here you have the solution: Integral of arctan(x)/(1+x)^2 dx = Parts: Integral of u dv = uv - Integral of v du u = arctan(x) ==> du = 1/(1+x^2) dx dv = 1/(1+x)^2 dx = (1+x)^(-2) dx ==> v = (1+x)^(-2+1)/(-2+1) = -1/(1+x) = arctan(x)*(-1/(1+x)) - Integral of (-1/(1+x))(1/(1+x^2)) dx = = -arctan(x)/(1+x) + Integral of 1/((1+x)(1+x^2)) dx = Parital fraction decomposition for 1/((1+x)(1+x^2)): 1/((1+x)(1+x^2)) = A/(1+x) + (Bx+C)/(1+x^2) = = A(1+x^2)/((1+x)(1+x^2)) + (Bx+C)(1+x)/((1+x)(1+x^2)) = = (A+Ax^2)/((1+x)(1+x^2)) + (Bx+C+Bx^2+Cx)/((1+x)(1+x^2)) = = (A+Ax^2+Bx+C+Bx^2+Cx)/((1+x)(1+x^2)) = = ( (A+B)x^2 + (B+C)x + (A+C) )/((1+x)(1+x^2)) 0 = A + B ==> A = -B 0 = B + C ==> C = -B 1 = A + C ==> 1 = -B + (-B) = -2B ==> B = -1/2 A = 1/2 C = 1/2 B = -1/2 1/((1+x)(1+x^2)) = A/(1+x) + (Bx+C)/(1+x^2) = (1/2)(1/(1+x)) - (1/2)((x-1)/(1+x^2)) = -arctan(x)/(1+x) + Integral of 1/((1+x)(1+x^2)) dx = = -arctan(x)/(1+x) + Integral of [ (1/2)(1/(1+x)) - (1/2)((x-1)/(1+x^2)) ] dx = = -arctan(x)/(1+x) + (1/2)Integral of 1/(1+x) dx - (1/2)Integral of x/(1+x^2) dx + (1/2)Integral of 1/(1+x^2) dx = = -arctan(x)/(1+x) + (1/2)Integral of 1/(1+x) dx - (1/4)Integral of 2x/(1+x^2) dx + (1/2)Integral of 1/(1+x^2) dx = = -arctan(x)/(1+x) + (1/2)ln|1+x| - (1/4)ln|1+x^2| + (1/2)arctan(x) + C ;-D

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[@IntegralsForYou]

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[@marcoovidiogautovega3222]

graciaaaaas tenia una integral parecida y no sabia como hacer hasta que vi el video

[@IntegralsForYou]

Me alegro mucho! Este canal cumple con su objetivo :D :D

[@aliali0001]

Last step??

[@IntegralsForYou]

sin(arctan(x)) = kzbin.info/www/bejne/d2TIept4obGrZrs 😉

[@SangramGaikwad-uv4hy]

Screen thode acchi dekhaya karo

[@IntegralsForYou]

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thanks that helped

[@IntegralsForYou]

You're welcome! 😊

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[@IntegralsForYou]

Thank you! ;-D

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tyty

[@IntegralsForYou]

😊😊

[@nagarjunagoudkokkisa4044]

integral (d^2+x^2)^(3/2) sir please explain this

[@IntegralsForYou]

Hi, Nagarjuna Goud Kokkisa! A long answer, but here it is: (I suppose d>0) Integral of (d^2 + x^2)^(3/2) dx = = Integral of [d^2(1 + (x/d)^2]^(3/2) dx = = (d^2)Integral of (1 + (x/d)^2)^(3/2) dx = Substitution: x/d = u (1/d) dx = du ==> dx = d du = (d^2)Integral of (1 + u^2)^(3/2) d du = = (d^3)Integral of (1 + u^2)^(3/2) du = Substitution: u = tan(t) ==> arctan(u) = t du = 1/cos^2(t) dt = (d^3)Integral of (1 + tan^2(t))^(3/2) 1/cos^2(t) dt = = (d^3)Integral of (1/cos^2(t))^(3/2) 1/cos^2(t) dt = = (d^3)Integral of (1/cos^3(t))*(1/cos^2(t)) dt = Parts: Integral of u dv = uv - Integral of v du u = 1/cos^3(t) ==> du = 3sin(t)cos^2(t)/cos^6(t) dt = 3sin(t)/cos^4(t) dt dv = 1/cos^2(t) dt ==> v = tan(t) = sin(t)/cos(t) = (d^3)[ (1/cos^3(t))(sin(t)/cos(t)) - Integral of (sin(t)/cos(t))*(3sin(t)/cos^4(t)) dt ] = = (d^3)[ sin(t)/cos^4(t) - 3*Integral of sin^2(t)/cos^5(t) dt ] = = (d^3)[ sin(t)/cos^4(t) - 3*Integral of (1-cos^2(t))/cos^5(t) dt ] = = (d^3)[ sin(t)/cos^4(t) - 3*Integral of 1/cos^5(t) dt + 3Integral of 1/cos^3(t) dt ]= = (d^3)[ sin(t)/cos^4(t) - 3*Integral of 1/cos^5(t) dt + 3* kzbin.info/www/bejne/h5vUmZWAd96Zgtk ]= = (d^3)[ sin(t)/cos^4(t) - 3*Integral of 1/cos^5(t) dt + 3*((1/2)sin(t)/cos^2(t) + ln|tan(t) + 1/cos(t)|) ] = (d^3)[ sin(t)/cos^4(t) - 3*Integral of 1/cos^5(t) dt + (3/2)sin(t)/cos^2(t) + 3*ln|tan(t) + 1/cos(t)| ] We have: Integral of 1/cos^5(t) dt = sin(t)/cos^4(t) - 3*Integral of 1/cos^5(t) dt + (3/2)sin(t)/cos^2(t) + 3*ln|tan(t) + 1/cos(t)| ==> Integral of 1/cos^5(t) dt + 3*Integral of 1/cos^5(t) dt = sin(t)/cos^4(t) + (3/2)sin(t)/cos^2(t) + 3*ln|tan(t) + 1/cos(t)| ==> 4*Integral of 1/cos^5(t) dt = sin(t)/cos^4(t) + (3/2)sin(t)/cos^2(t) + 3*ln|tan(t) + 1/cos(t)| ==> Integral of 1/cos^5(t) dt = (1/4)[ sin(t)/cos^4(t) + (3/2)sin(t)/cos^2(t) + 3*ln|tan(t) + 1/cos(t)| ] ==> Integral of 1/cos^5(t) dt = (1/4)sin(t)/cos^4(t) + (3/8)sin(t)/cos^2(t) + (3/4)*ln|tan(t) + 1/cos(t)| We had: Integral of (d^2 + x^2)^(3/2) dx = = ... = = (d^3)Integral of (1/cos^3(t))*(1/cos^2(t)) dt = = Integral of 1/cos^5(t) dt = = (1/4)sin(t)/cos^4(t) + (3/8)sin(t)/cos^2(t) + (3/4)*ln|tan(t) + 1/cos(t)| = = (1/4)sin(arctan(u))/cos^4(arctan(u)) + (3/8)sin(arctan(u))/cos^2(arctan(u)) + (3/4)*ln|tan(arctan(u)) + 1/cos(tarctan(u))| = = (1/4)sin(arctan(x/d))/cos^4(arctan(x/d)) + (3/8)sin(arctan(x/d))/cos^2(arctan(x/d)) + (3/4)*ln|tan(arctan(x/d)) + 1/cos(tarctan(x/d))| + C It can be simplified but this answer should be correct if I didn't do any mistake...

[@nagarjunagoudkokkisa4044]

Tq sir

[@IntegralsForYou]

You're welcome! ;-D

[@SOLVINGMATHPROBLEMS]

Unnecessarily made lengthy.

[@NehaSingh-xb3cw]

integration x^13/2.(1+x^5/2)^1/2 please help me in solving dis question

[@IntegralsForYou]

Hi Bikesh Singh! Here it is the answer (write it on the paper and you will see it better than here on the screen...): Integral of x^(13/2)*(1+x^(5/2))^(1/2) dx = = Integral of x^(5/2)x^(5/2)x^(3/2)*(1+x^(5/2))^(1/2) dx = = Integral of x^(5/2)x^(5/2)*(1+x^(5/2))^(1/2) x^(3/2)dx = Substitution: u= (1+x^(5/2))^(1/2)==> u^2=1+ x^(5/2) ==> u^2 - 1 = x^(5/2) du = (1/2)(5/2)x^(3/2)(1+x^(5/2))^(-1/2) dx ==> (2/1)(2/5)(1+x^(5/2))^(1/2) du = x^(3/2) dx ==> ==> (4/5)u du = x^(3/2) dx = Integral of (u^2-1)(u^2-1)*u (4/5)u du = = (4/5)Integral of (u^2)(u^2-1)^2 du = = (4/5)Integral of (u^2)(u^4 - 2u^2 + 1) du = = (4/5)Integral of (u^6 - 2u^4 + u^2) du = = (4/5)( (1/7)u^7 - (2/5)u^5 + (1/3)u^3) ) = = (4/5)( (1/7)((1+x^(5/2))^(1/2))^7 - (2/5)((1+x^(5/2))^(1/2))^5 + (1/3)((1+x^(5/2))^(1/2))^3) ) = = (4/5)( (1/7)(1+x^(5/2))^(7/2) - (2/5)(1+x^(5/2))^(5/2) + (1/3)(1+x^(5/2))^(3/2) ) + C You can simplify as you want. For example when we have: (4/5)( (1/7)u^7 - (2/5)u^5 + (1/3)u^3) ) = (4/5)(u^3)( (1/7)u^4 - (2/5)u^2 + (1/3)) ) :-D

[@mukulrohilla3894]

Can we solve this integration without using trigonometric fn? And if (1+x^2)^3/2 replace by (z^2+x^2)^3/2....??

[@IntegralsForYou]

Hi Mukul Rohilla! I don't think we can do it without using trig substitution... Maybe you can try doing u=1+x^2 or u=(1+x^2)^(1/2) but I don't think it will work, and if so, I think you will have to a trig substitution after the first substitution. If we have (z^2+x^2)^(3/2) then you have to do u = z*tan(x). The final solution will be z^2/(z^2+x^2)^(1/2) + C.

[@mukulrohilla3894]

But in a book solution is x/z^2(z^2+x^2)^1/2. what should I do?

[@IntegralsForYou]

Sorry Mukul Rohilla, I did a mistake writing you the answer, the final solution is x/( (z^2)(z^2+x^2)^(1/2) ) as you said: Integral of 1/(z^2+x^2)^(3/2) dx = Substitution: x = z*tan(u) ==> x/z = tan(u) => arctan(x/z) = u dx = z/cos^2(u) du = Integral of 1/(z^2 + (z^2)tan^2(u))^(3/2) z/cos^2(u) du = = Integral of 1/( (z^2)(1 + tan^2(u)))^(3/2) z/cos^2(u) du = = Integral of 1/( (z^2)^(3/2)(1 + tan^2(u))^(3/2) z/cos^2(u) du = = Integral of 1/( (z^3)(1/cos^2(u))^(3/2) z/cos^2(u) du = = Integral of 1/( (z^3)(1/cos^3(u)) z/cos^2(u) du = = Integral of z*cos^3(u)/(z^3)cos^2(u) du = = Integral of cos(u)/z^2 du = = (1/z^2)Integral of cos(u) du = = (1/z^2)sin(u) = = (1/z^2)sin(arctan(x/z)) = = (1/z^2)((x/z)/sqrt(1 + (x/z)^2)) = = (1/z^2)(x/z*sqrt(1 + x^2/z^2)) = = (1/z^2)(x/z*sqrt(z^2/z^2 + x^2/z^2)) = = (1/z^2)(x/z*sqrt((z^2+x^2)/z^2)) = = (1/z^2)(x/(z/z)sqrt(z^2+x^2)) = = (1/z^2)(x/sqrt(z^2+x^2)) = = x/(z^2)*sqrt(z^2+x^2) + C

[@mukulrohilla3894]

Thank you

[@iqrazahid8116]

One more help plz if we put 3 instead of 1 in qs?

[@IntegralsForYou]

(3+x^2)^(3/2) = ( 3*(1+(x^2)/3)^(3/2) = (3^(3/2))*(1 + (x/sqrt(3))^2)^(3/2) = 3*sqrt(3)*(1 + (x/sqrt(3))^2)^(3/2) Integral of 1/(1+x^2)^(3/2) dx = = Integral of 1/[ 3*sqrt(3)*(1 + (x/sqrt(3))^2)^(3/2) ] dx = Substitution: t = x/sqrt(3) dt = 1/sqrt(3) dx ==> sqrt(3)dt = dx = Integral of 1/[ 3*sqrt(3)*(1+t^2)^(3/2) ] sqrt(3)dt = = (1/3)*Integral of 1/(1+t^2)^(3/2) dt = = (1/3)* kzbin.info/www/bejne/hqC6mYV3apaKaM0 = = (1/3)*(t/sqrt(1+t^2)) = = (1/3)*( (x/sqrt(3))/sqrt(1+(x/sqrt(3))^2) = = x/[ 3*sqrt(3)*sqrt(1+(x/sqrt(3))^2) ] = = x/[ 3*sqrt( 3*(1+(x^2)/3) ) ] = x/[3*sqrt(3+x^2)] + C ;-D

[@jyotitiwari2138]

Sr i have a ques

[@phuongpham1549]

thanks so much love love

[@IntegralsForYou]

You're welcome! ;-D

[@abdulhananfarooq8515]

why are we putting x equal to tan(u)????

[@IntegralsForYou]

Hi! The general rule for trig substitution is: If sqrt(a^2 - x^2) ==> x = a*sin(u) or x = a*cos(u) ==> x/a = sin(u) or x/a = cos(u) If sqrt(a^2 + x^2) ==> x = a*tan(u) ==> x/a = tan(u) If sqrt(x^2 - a^2) ==> x = a*sec(u) ==> x/a = cos(u) In this case, we have (1+x^2)^(3/2) = (1+x^2)(1+x^2)^(1/2) = (1+x^2)sqrt(1+x^2) . Since we have a sqrt(1+x^2) we try x=tan(u) and it works ;-D

[@the_abhinav0806]

This is to long solution of this question.... I will solve it in just four lines..

[@IntegralsForYou]

If I want to explain all the details I have to do it long but if you can skip some steps, then do it! 💪😊

[@pratyush7987]

You can give a solution for this in 2 lines if you skip steps but he gave an elaborated explanation to help beginners.

[@umasarkardey7638]

​@@pratyush79871:03

[@iamimran5880]

Thank you sir

[@IntegralsForYou]

You're welcome! ;-D

[@iamimran5880]

@@IntegralsForYou sir I want a solution of a question

[@iamimran5880]

@@IntegralsForYou integration of (1-x^4)^3/2÷(1+×^4)^2 limit 0 (lower) to 1 (upper) please help me

[@IntegralsForYou]

Hi, Imran ahmad, I don't think it can be done by getting the indefinite integral, because it cannot be expressed in terms of elementary functions. Are you sure you aren't asked to use numerical methods?

[@johnr1875]

yeah, how did you do that last part?

[@IntegralsForYou]

Hi John Rogers, if you are asking for the sin(arctan(x)) part, it is done in this video: kzbin.info/www/bejne/d2TIept4obGrZrs

[@johnr1875]

Integrals ForYou thank you

[@IntegralsForYou]

;-D

[@johnr1875]

Integrals ForYou my guy, you should seriously make more videos. You really are the man.

[@IntegralsForYou]

hehe I still have a lot of integrals in my TODO list ;-D Thanks for your comment!

[@nishachoudhary9952]

Ye aap likh kaise rahe ho

[@Blackberry405]

I mean I understand. But wtf !!

[@IntegralsForYou]

😜

[@Unknown_31415]

bro I fell you

[@elizabethpeacock9054]

What level of calculus is this? College level Calc 1or 2?

[@dhruv7915]

Beth Peacock thats high school calculus here in india

[@arianas7866]

@@dhruv7915 omg plz not this again. I've seen a million of these comments before Yes it is not only in india they teach calculus we do it here in the UK at high school too surprise surprise

[@dhruv7915]

@@arianas7866 cool

[@tanusrisantra7111]

Sounds kha h

[@dilanramoscalizaya2207]

👍👏

[@IntegralsForYou]

😊

[@stipepavic843]

thxxxxxxxxxx

[@IntegralsForYou]

:-D

[@shashankkumar2032]

Sir write clearl

[@sccm100]

holy fuck

[@robertnowak5477]

yo should be (-3/2)

[@IntegralsForYou]

Hi Robert! Where should it be -3/2 ? Thanks!

[@robertnowak5477]

@@IntegralsForYou Nvm haha sry

[@IntegralsForYou]

No problem! 😉

[@priyanshugupta7905]

Bhai mhuuu se kuch boll

[@IntegralsForYou]

Hi, I know that it would be easier for you if I talked but I prefer to let you think why am I doing each step and if you don't understand it you can ask me in a comment. In my opinion, it is the best way to learn to integrate 😉

[@priyanshugupta7905]

@@IntegralsForYou great bro

[@IntegralsForYou]

@priyanshu gupta Thanks! 💪

[@kiit8337]

@@IntegralsForYou how did u get him 😂😂

[@kdh00961]

빡통대가리의 구원자 센세 아리가또.....

[@AdibaAhmed-s2p]

Handwriting is so bad

[@IntegralsForYou]

🤷

[@xiaodongyu2105]

TYTY

[@IntegralsForYou]

:d :d

[@世界哥看世界]

wrong answer should be x ( 2 x 2 + 3 ) 3 ( x 2 + 1 ) 3 2 + C

[@IntegralsForYou]

Hi! I am sorry but I don't understand your answer