(Method 1) Integral of sqrt(1 + sin(x)) (substitution)

[@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/

[@alankuo5579]

You can use sinx=2sin(x/2)cos(x/2) and then 1=sin(x/2)^2+cos(x/2)^2 and KaBOOM you have (sin(x/2)+cos(x/2))^2

[@IntegralsForYou]

Hi! I did something similar here kzbin.info/www/bejne/lWa6kJmmatxlrNU

[@rasmusturkka480]

Omg, I'm so dumb, I thought about that sub but didn't try and see what happens

[@therawmusician2833]

AN IDEAL SOLUTION OF ANY QUESTION IS THE EASIEST ONE. Easy MEthod- put 1=sinsquarex/2+cossquarex/2 & sinx=2sinx/2cosx/2 {BY HALF ANGLE FORMULA} Then it will be like INTEGRATIONroot ( sinsquarex/2+cossquarex/2+2sinx/2cosx/2) USE THE IDENTITY ---- { Asquare+Bsquare+2AB = (A+B)whole square } implies that -- INTEGRATIONroot{(sinx/2+cosx/2)wholesquare} [ ROOT CANCELS WHOLE SQUARE ] implies that -- INTEGRATION(sinx/2+cosx/2) = -2cosx/2+2sinx/2 + C = 2(sinx/2-cosx/2) + C ANSWER...............................

[@IntegralsForYou]

Hi! I did it here ==> kzbin.info/www/bejne/lWa6kJmmatxlrNU 😉

[@JBChem]

Damn dude you made it whole lot easier Thanks!

[@mbond5358]

I like both methods.

[@SMILEARABIC]

I can only see a giant hand

[@teriatilatifa5401]

روح الله يعطيك الربح والخير ويهنيك وين ما تروح 💜♥️♥️

[@IntegralsForYou]

Thank you! 😊😊

[@STKeTcH]

The man, the myth, the mathlegend!

[@IntegralsForYou]

😊

[@udaysharma3646]

Thanks🙏🙏🙏🙏 From 🇮🇳 India

[@udaysharma3646]

I saw your solution only upto integral of √cos^2x/√1-six dx then I got the idea. It better to practice on own but if you are facing some difficulty then you can see 1-2 steps and then try on your own.

[@IntegralsForYou]

@Uday Sharma Totally agree! Thank you for your comments! 😊

[@TONIO-ru4iu]

I've succeeded in integrating function sqrt(1+sin(x)) by using trigonometric identities!!! The answer is sgn(cos(x)+sin(x))*(sin(x)-cos(x))!!!

[@ROHIT_kumar_999]

Thanks from 🇮🇳

[@IntegralsForYou]

My pleasure! ❤

[@vietle7340]

how you can calculate the definite integral from 0 to 2pi, if we use sub, we get u=1 for both x.

[@tabiematics]

Hello, you are missing the absolute value for cosx , because of taking the square root of t^2 is |t| , not t since t is negative , on the other hand, you can check your answer, if you integral from 0 to 2Pi , then you can get non-zero at left hand side and get zero at right hand side .

[@digantamajumder5900]

In a question, they put a limit from 0 to π. From the integral value the answer is zero. But in other processs (making a whole square form of 1+sinx) the answer is 4. Please help!

[@arishmitghosh3016]

you can replace 1 by sin^2 (x/2) + cos^2(x/2) and you can write sin x as (2sin(x/2).cos(x/2)) thus it becomes (sin(x/2)+cos(x/2))^2

[@meenakshisaju9482]

I have the same doubt too. If we use both the methods, is the answer the same? I too am getting 0 but the actual answer is 4.

[@danielliu6952]

This is a late reply, but I believe when the sqrt(sin^2(x)) happens, the value is not actually sin(x), but the absolute value of sin(x)

[@digantamajumder5900]

@@danielliu6952, you are too late dude. I haven't touched mathematics since 2020 😂... But can you elaborate your opinion little bit?

[@mbond5358]

Bravo, Note: -2u+C. sin(x) not equal 1. It is very important for definite integration(s) .

[@IntegralsForYou]

Thanks again, M Bond! 😊

[@IrisIris-jl9zy]

Thank you mate 🔥

[@IntegralsForYou]

My pleasure! ❤ Please help me grow my new channel about derivatives by subscribing 👉 www.youtube.com/@DerivativesForYou?sub_confirmation=1 Thank you! ❤

[@sparshnagpal1509]

But graphically won't it be positive? Integral something in root should always be plotted in +be y region so the Shouldn't value of area found by it also be positive?

[@sakettiwari7715]

I-sinx=t*2 Is also a good substitute

[@IntegralsForYou]

Yes, in fact, it is the same substitution: u = sqrt(1-sin(x)) ==> u^2 = 1-sin(x)

[@sakettiwari7715]

Yes sir

[@IntegralsForYou]

@Saket Tiwari 😉

[@lokeswarimarrapu2222]

Super exaplination

[@IntegralsForYou]

Thanks! 💪

[@giuseppedarro4766]

what if its`s the reciprocal? 1/sqrt(1+sin(x)) ????

[@bhavanaverma8553]

Dude you are a genius 💜🙆

[@IntegralsForYou]

Thanks, Bhavana Verma! ;-D

[@saran9736]

Hey, why not convert the sinx as cosx using T ratio conversion and then apply submultiple angle formula, I got -2√2sin(π/4 - X/2)+C as the answer

[@IntegralsForYou]

Hi! Of course you can use it! And I did it in another way too here kzbin.info/www/bejne/lWa6kJmmatxlrNU . I wrote it all in terms of x/2 and then simplify the square root 😉

[@saran9736]

@@IntegralsForYou Ah yes, thanks, hey by the way i wonder whether I can convert this sinx as cosx and then proceed the answer is ever so slightly different to this one, I guess that way is correct too!?

[@IntegralsForYou]

@@saran9736 hey, I'm sorry, I don't understand your question...

[@mbond5358]

@Saran, your method also works!

[@williamcolgate5708]

How many of you were solving Cengage?

[@IntegralsForYou]

I didn't know about Cengage 😂😞

[@Sam1-t7q]

2:46 how did you write there "dx" and not "dx/du"

[@IntegralsForYou]

Hi! Because when use the substitution method we derive in both sides and then we add "dx" in the "x" part and "du" in the "u" part of the expression

[@galaxyman195]

What integral should I use to solve integral 1/sqrt(x+sinx) dx for comparison test

[@IntegralsForYou]

Hi! Without knowing the integration limits you have I think you are looking for the function f(x)=1 because 1/sqrt(x+sinx) ≤ 1

[@a.m.l.7004]

Incorrect. The answer is correct iff 1-sin(x) is different from 0 (x is different from pi/2)

[@TheNikhilify]

My answer by a different method is -2cosx/2 + 2sinx/2 + C

[@digantamajumder5900]

This the whole square form. It's same.

[@Gagan-sf5gy]

How bro tell me your method

[@khalafibrahem3928]

Integral from (π) to (-π) Sqrt(1+sinx)

[@黃詩伃-r8j]

Why is the √cos^2 is cos not -cos

[@Kenzo-0-]

Dude u r awesome, ty for videos

[@IntegralsForYou]

You're welcome Madara, thanks for your comment!

[@sakibsonnet2789]

Thank you. :-)

[@IntegralsForYou]

You're welcome :D

[@ahmedsakr7497]

integration sin2x/√1+sinx

[@manishkumarsingh3082]

I used 1+sin(x)=cos^2(x/2)+sin^2(x/2)+2sin(x/2)cos(x/2) So sqrt will open with mod(cos(x/2)+sin(x/2) or not??

[@TheNikhilify]

Yes

[@dailyenglishepisodes]

Hey bro! What are you doing??? It's simple...Why are you making it complicated??

[@IntegralsForYou]

Hi AL MAMUN! Am I making it so complcated? It is just a method, check out the method 2! ;-D

[@abhichaturvedi8322]

You are good ,, please try to add sound

[@IntegralsForYou]

I'll try! Thanks for your comment!

[@mavi6242]

Abi çok iyi ya

[@IntegralsForYou]

😊

[@ctetnewsclub1752]

Arey bhai sound kaha hai

[@HydrodynamicModel]

sqrt(Cos^2(x))=Cosx ?????????????????? No

[@Inndifference]

I have the same exact question just the sinx is squared please help me.

[@IntegralsForYou]

Hi, Indifference, I need more details to help yopu

[@Inndifference]

Integral of sinx^3 cosx square root of 1+sinx^2 dx

[@Inndifference]

I tried table of integrals but apparently it didn’t work.

[@Inndifference]

Oh my god thanks so much you saved my life. You are my new friend:)

[@IntegralsForYou]

You're welcome!! 😀

[@chiku918]

Why i are not speaking not understandable

[@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 practice integration 😊

[@chiku918]

Please solve it Integration root 1+sin dx = Option a = 1/2(sinX/2+cos x/2) b= 1/2(sinx/2-cos x/2) c=2 root 1+sinx +c d= - 2 root 1-sinx+c Bro solve it

[@IntegralsForYou]

@@chiku918 Hi! I did this integral in two different ways: Method 1: integralsforyou.com/integrals?v=h7V-Od0nloc Method 2: integralsforyou.com/integrals?v=_5W_fp5w2vs Let me know what do you think 🙂

[@chiku918]

@@IntegralsForYou Brother your answers is not in my options

[@IntegralsForYou]

@@chiku918 - Integral of sqrt(1+sin(x)) dx = integralsforyou.com/integrals?v=h7V-Od0nloc = -2sqrt(1-sin(x)) + C (option D) - Integral of sqrt(1+sin(x)) dx = integralsforyou.com/integrals?v=_5W_fp5w2vs = (1/2)(sin(x/2) - cos(x/2)) + C (option B) Hope it helped! 💪

[@MustafaAli-eh6zq]

how can we get (-cosx) values in du

[@IntegralsForYou]

Hi Mustafa ali! Here we apply the chaine rule for the derivative of sqrt(1-sin(x)). Derivative of sqrt(x) = 1/2sqrt(x) Derivative of 1-sin(x) = 0 - cos(x) = -cos(x) So: Derivative of sqrt(1-sin(x)) = [1/2sqrt(1-sin(x))](-cos(x)) = - cos(x)/2sqrt(1-sin(x)) And: du = - cos(x)/2sqrt(1-sin(x)) dx -2du = cos(x)/2sqrt(1-sin(x)) dx Hope you understand now :D

[@wahyuauliaakbar7914]

@@IntegralsForYou I think its true

[@wahyuauliaakbar7914]

And thanks for your ideas

[@IntegralsForYou]

@@wahyuauliaakbar7914 Hi! Since the derivative of sin(x) is cos(x), then the derivative of -sin(x) is -cos(x). We have also that the derivative of a constant is zero, then the derivative of 1 is zero. Finally: Derivative of 1-sin(x) = 0 - cos(x) = -cos(x)

[@wahyuauliaakbar7914]

@@IntegralsForYou sorry I forget it.

[@kishorekumarsathishkumar1562]

Dude, just use 2 u-subs and you get the same thing...

[@IntegralsForYou]

Hi! Which substitutions you would do? Thanks!

[@ashupatelpatel2394]

B

[@rapgod8710]

Integrate square root (1+sinx/2)

[@IntegralsForYou]

sin(2t) = 2sin(t)cos(t) ==== t=x/4 ====> sin(x/2) = 2sin(x/4)cos(x/4) 1=sin^2(t)+cos^2(t) ==== t=x/4 ====> 1=sin^2(x/4)+cos^2(x/4) a^2+b^2+2ab = (a+b)^2 Integral of sqrt(1+sin(x/2)) dx = = Integral of sqrt( sin^2(x/4) + cos^2(x/4) + 2sin(x/4)cos(x/4) ) dx = = Integral of sqrt( (sin(x/4) + cos(x/4))^2 ) dx = = Integral of ( sin(x/4) + cos(x/4) ) dx = Substitution: u = x/4 du = (1/4) dx ==> 4 du = dx = Integral of ( sin(u) + cos(u) ) 4 du = = 4*Integral of ( sin(u) + cos(u) ) du = = 4*( -cos(u) + sin(u) ) = = 4*( sin(u) - cos(u) ) = = 4*( sin(x/4) - cos(x/4) ) + C

[@Nmhuy321]

Lul i first thought it was ln(sinx)

[@lmene9720]

Thank so much you are a gorgeous 🤩🤩

[@IntegralsForYou]

Thanks, racha faracha! 😉😉

[@mbs0710]

Wow

[@IntegralsForYou]

;-D

[@ShivamMishra-fd9dl]

Integrate root of 1 + sin(x/n)dx

[@IntegralsForYou]

Hi Shivam Mishra! You only have to do a substitution an then you will have the integral of sqrt(1+sin(x)) Integral of sqrt(1+sin(x/n)) dx = Substitution: u = x/n du = 1/n dx ==> n du = dx = Integral of sqrt(1+sin(u)) n du = = n*Integral of sqrt(1+sin(u)) du = = ... and then you do the same as I do in the video ;-D

[@devbratnaik5399]

I don't like your method

[@IntegralsForYou]

I have two methods and you just watched the Method 1. What do you think about the Method 2? 😉 Method 1: kzbin.info/www/bejne/nmi5XoKaZdOfpcU Method 2: kzbin.info/www/bejne/lWa6kJmmatxlrNU

[@pradeepkathait1149]

wt a nonsense!!