* Simplify the expression: * We can use the identities tan(x) = sin(x) / cos(x) and cosec(x) = 1 / sin(x) to rewrite the expression: sin(x) tan(x) + cos(x) cosec(x) = sin(x) * (sin(x) / cos(x)) + cos(x) * (1 / sin(x)) = sin^2(x) / cos(x) + cos(x) / sin(x) * Integrate using the formulas: * Now we can integrate each term separately using the formulas for sin(x) and cos(x): ∫ (sin^2(x) / cos(x) + cos(x) / sin(x)) dx = -∫ cos(x) dx + ∫ sin(x) dx = sin(x) - cos(x) + C Therefore, the integral of sin x tan x + cos x cosec x is sin(x) - cos(x) + C Bohot Jada Mehnat Lagi hai type karne me 😅
Change tan x into sinx/cosx and cosec x as 1/sinx then ; It becomes sin^2x / cos x + cosx /sinx then: Write sin^2x as 1-cos^2x then seprate and solve and ans will be ln|secx+tanx| + ln|sinx| - sinx + c ❤
@@prt-tanish5547bro please guide me i am new in class 11th. Which coaching should I take How should I start preparation and which books to follow? How did you prepared. Which mistakes I don't have to do
@@UdaySingh-rn4fx mathematics me ek X ki equation ke liye infinite number of values aa jati hai to ye to algebric me hai Phir bhi kaise nikala bta deta hu Sin.tan ko sin²/cos banaya aur cosec.cos ko cos/sin banaya Phir sin²/cos me sin² ko 1 - cos² se replace kiya aur cos/sin me let sin=t krke sovle kiya aur answer agya Log|sec + tan| - sin + log|sin| + c
@@ratnajain-bs8dzbhai tum online padhte ho ya offline? Mai bhi gb sir se integration padh rha hu telegram ke lectures se and cengage illustration solve kar rha hu
Bruhh Pehle stanx ko sinx/cosx likho aur for sin¢2x ko 1-cos^2x likhdo jisse eqn bnjaaegi integration of secx+cosx Second part ke lie cot theta ka integration krdo Basic formulaes Ez maths
Uss sawaal me direct product nikalo aur sin square theta ko 1-cos square theta kardo Aur dusre plus wale part me niche wale cos square theta se cos theta =t lelo phir go upar uska derivative hai phir game jaisa khelke kaam tamam kardo
§(tanx.sinx+cosecx.cosx)dx By using formula tanx=sinx/cosx Cosex=1/sinx So we can write it §(sinx. sinx /cosx+cosx .1/sinx We know that cosx/sinx=cotx and sin^2x=1-cos^2x So §(1-cos^2x/cosx+cotx)dx §secxdx-§cosxdx+§cotxdx ln|secx+tanx|+sinx+ln|sinx|+c😅
Limit not difined hai answer alag ho sakta hai Acording to me- Int(sinxtanx)=> Int(sin²x/cosx)=> Int(1-cos²x)/cosx=> Int(1/cosx - cos²x/cosx)=> Int(secx-cosx) Int(secx) - int(cos)=> Ln|tanx +secx| - sinx+c Similarly 2nd part
So answer is firsty do sepret devaid=√sinxtanxdx+√cosecxcosx and weknow sinx=cosx/cotx and tanx =sinx/cosx put in √sinxtanx then we have = √sinx/cotxdx and sinx= cosx/cotx and put in after this we have *√cosx* then we go secand part √cosecxcosxdx we know cosex =cotx/cosx put and we have *√cotx* after all thi we have √cosxdx+√cotxdx after integret this we have fainali*sinx+log[six]+c* Ungliyo me dard ho gya 😂😂😂😊
mera answer root log(cosx) kyu a ra hai? btw i just started integration today, (2026tard) ky ham values ko root me calc kr skte he kya? like my thing came out to be integral of root (sec x d(cosx))
From chat gpt To solve the expression \(\sin\theta \tan\theta + \csc\theta \cos\theta\), let's break it down into simpler parts using trigonometric identities. 1. \(\tan\theta = \frac{\sin\theta}{\cos\theta}\) 2. \(\csc\theta = \frac{1}{\sin\theta}\) Using these identities, the expression becomes: \[ \sin\theta \left(\frac{\sin\theta}{\cos\theta} ight) + \left(\frac{1}{\sin\theta} ight) \cos\theta \] Simplify each term separately: \[ \frac{\sin^2\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} \] Combine the fractions over a common denominator: \[ \frac{\sin^2\theta \sin\theta + \cos^2\theta \cos\theta}{\cos\theta \sin\theta} \] Simplify the numerator: \[ \frac{\sin^3\theta + \cos^3\theta}{\cos\theta \sin\theta} \] Since the numerator cannot be simplified further with common trigonometric identities, the expression \(\sin\theta \tan\theta + \csc\theta \cos\theta\) simplifies to: \[ \frac{\sin^3\theta + \cos^3\theta}{\cos\theta \sin\theta} \] Therefore, the simplified form of the given expression is: \[ \frac{\sin^3\theta + \cos^3\theta}{\cos\theta \sin\theta} \]
To integrate the given expression, we can start by rewriting it in a more tractable form: sin(theta)tan(theta) + cosec(theta)cos(theta) = sin(theta)(sin(theta)/cos(theta)) + (1/sin(theta))(cos(theta)) = (sin^2(theta) + cos(theta))/cos(theta) = (1 + cos(theta))/cos(theta) (using the Pythagorean identity sin^2(theta) + cos^2(theta) = 1) = sec(theta) + 1 Now, we can integrate: ∫(sec(theta) + 1) dtheta = ∫(1/cos(theta)) dtheta + ∫(1) dtheta = ln|sec(theta) + tan(theta)| + theta + C where C is the constant of integration. Therefore, the integral of sin(theta)tan(theta) + cosec(theta)cos(theta) is ln|sec(theta) + tan(theta)| + theta + C.
Mene socha easy hai karte hai phir dekha to 1 ghanta lag gya phir jab dobara short dekha to pata chala vo second part cosec cos tha mene cotcosec le liya
Separate the integral sin theta*tan theta and cosec theta*cos theta And solve separately by integration by parts🤗method its so easy😅,you won't get trapped in this part ANSWER IS ln(secx+tanx)+ln(sinx)-sinx +c and i counted 1 as a constant so i ignored it,replace x with theta🤓
Solution: 1.Write the expression in terms of sinx and cosx. 2.Convert all sinx into cosx using [(sinx)^2 +(cosx)^2 =1] 3.Divide and break the fraction (a-b/c =a/c-b/c) Remaining: secx-cosx+cosecx.cotx Now integrate... Answer: ln(|secx+tanx|) - sinx - cosecx + c
Bhai phele wala simply hojayega jab tan theta ko sinx/cosx likhoge upar 1-cos²x bach jayega fir numerator ko alag alag krlo Aajyega pehle wle ka ans Aur dusra wala part to cot theta likha hua h jiska integration baccha baccha janta h lnlsinxl Btw and is Lnlsecx+tanxl + sinx + lnlsinxl + C C is very important btw 😅😅
Simplifying the expression =int(sin²x/cosx + cosx/sinx)dx =int(1-cos²x/cox + cotx)dx =int(1/cosx-cos²x/cosx)dx + int(cotx)dx =int(secx-cosx) + log(sinx) + C =log(secx+tanx) -(-sinx) + log(sinx) +C And the final answer is =log{(sinx+sin²x)\cosx} + Sinx + C
Bhai pehle dono terms ko alag kr lo second term simple hai ... Second term ko simplify kroge to cot x aa jayega jiska intigration log |sinx|+ C or Rahi baat first term ki to tan ki din cos me change kr lo aur sin ² upper cos niche phir sin ko 1-cos²x likho phr term aa jayega 1-cos²x / cos x . Ab dono ko alag kr lo 1/cos x = secx aur cos ²x / cos x = cos x aayega. Ab dono ko alag alag intigrate krenge cos ka sin aur sec ka log |secx + tanx|. Final answer : log|sinx|- sinx+log|secx +tanx| + C
Bhaiya aap iss question ko aise solve kijiye tantheta ko sin/cos break karlo aur likh do sinsquare theta /costheta then sintheta ko t man lo phir aapka expression t^2dt aa jayega phir udhar aap cosec theta ko 1/sintheta kardo wo banjayega cot theta aur ha pehle integration ko break karlena aapka answer phir ayega sin^3theta/3+ln sintheta +c 😊
Use by part Best method to solve integration is remember 15. Ways or type of inte. Method which ur good teacher gave u in notes Aur xam m 1 min phir last m inte ka ques ko time deena
First rule of integration: kabhi bhi koi bhi question ese hi shuru na kardo pehle uski validation Karo , tabhi shuru Karo , ho sakta hai wo non integrable ho... *Agar sawaal na aaye to ye gyaan pel ke patli gali se nikal lo.😂
Question incomplete hai integral ka sign bas sin theta + cos theta ke aage laga hai😂😂 Kiske respect mai karna hai wo bhi nahi pata 😂 Bonus milna chahiye bhai ko💀💀💀💀
The correct answer to this question is ln(tanx+secx)+ln(sinx)-sinx+C we are taking theta is equal to x here since I don't know how to write theta in keyboard
For sinx tanx = break in sin cos, 1-cos sqaure/ sin , take sin t and alag alg solve khatam Now the next funtion break in sin cos, sin se multiply upr niche, 2sixcos=dt and sin sqaure x=t Khatam
Sir woh 2nd part easily ho jata hai.... Sirf first part ko sin and cos ke terms me likho aur fir take cos as t fir integration by parts... fir (1-t²) ko v manlo
To integrate the expression sin(x)tan(x) + cos(x)cosec(x), we can use the following trigonometric identities: - tan(x) = sin(x)/cos(x) - cosec(x) = 1/sin(x) Using these identities, we can rewrite the expression as: sin(x)tan(x) + cos(x)cosec(x) = sin(x)(sin(x)/cos(x)) + cos(x)(1/sin(x)) Simplifying, we get: = sin^2(x)/cos(x) + cos(x)/sin(x) Now, we can integrate term by term: ∫(sin^2(x)/cos(x) + cos(x)/sin(x)) dx = ∫(sin^2(x)/cos(x)) dx + ∫(cos(x)/sin(x)) dx Using substitution and integration by parts, we get: = -sin(x) + ∫(1/sin(x)) dx = -sin(x) - ∫(1/sin(x)) dx Now, using the identity ∫(1/sin(x)) dx = log|tan(x/2)| + C, we get: = -sin(x) - log|tan(x/2)| + C where C is the constant of integration. So, the final answer is: -sin(x) - log|tan(x/2)| + C
