Integral of cos 2x dx

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Integral of cos 2x dx

Home ? ?Mathematics? Mathematics Solution ?Integral Celle Problem? Previous Article Close Articles (Last updating on: January 19, 2020) Declaration of the problem: CE Board May 1995 What is the integral of Cos 2x and ^ (SIN 2x) DX? Problem response: The integral of the trigonometic function is (1/2) (E1 (SIN 2x)) + C. Saturdays of integral calculation problems more questions In: Integral Celle Questions and Answers Online in the Integral Celle ANTERT Next ARTICLE COS 2x is not the same as the COS2X integral. But by finding the integral of COS2x, we use the integral of Cos 2x too. To find this, we use the 2x fan and trigonomic identities. We use the replacement method to find these integral. We will find the integral of COS 2x and the integral of the COS2x and we also solve some problems related to these integral. What is the integral of Cos 2x DX? The integral of COS 2x DX is denoted by ? ?Cos 2x DX and its value is (SIN 2X) / 2 + C, where 'C' is the integration constant. To prove this, we use the replacement method. For this, assume that 2x = u. Then 2 dx = du (or) dx = du / 2. Replacing these values in the integral ? ?Cos 2x dx, cos 2x dx =? Cos u (du / 2) = (1/2) because you You know that Cos X integral is sin x + c. Thus, = (1/2) Sin U + C Replacing U = 2x Behind, ?Cos 2x DX = (1/2) Sin (2x) + C This is the integral of Cos 2x Formula. Defined Integral of COS 2x A defined integral is not all of a whole of an integral with lower and higher limits. By the fundamental calculation theorem, to calculate a definitive integral, we replace the upper limit first, and then the lower limit in the integral and subtract them in the same order. In this process, we can ignore the integration constant. Let's calculate some defined integral of the integral cos 2x dx here. Integral of COS 2x from 0 to 2ps \ '\ (?l ^ {2 \ pi} \) COS 2x dx = (1/2) Sin (2x) \ (\ Left. \ Dirtion | _0 ^ {2 \ pi} \) = (1/2) Sin 2 (2nd ) - (1/2) Sin 2 (0) = (1/2) Sin 4? ?g - Sin 0 = (1/2) 0 - 0 = 0, therefore, the integral of COS 2x of 0 to 2pi is 0. Cos 2x 2x of 0 to the PI ? \ \ (\ \ \ \) cos 2x dx = (1/2) SIN ( 2x) \ (\ left. \ Right | _0 ^ {\ pi} \) = (1/2) Sin 2 (?nsim ) - Sin 2 (0) = (1/2) Sin 2 ? - Sin 0 = 0 - 0 = 0, therefore, the integral of Cos 2x of 0 to pi is 0. What is the integral of cos ^ 2x dx? Cos2x's integral is denoted by ? ?Cos2x DX and its value is (x / 2) + (SIN 2x) / 4 + C. We can prove it in the two methods to follow. Using Formula 2x Using Integration by the Method of Pieces 1: COS 2X Integration Using Double Angle Formula To find the integral COS2x, we use the Double Ngle of Cos. One of the 2x cosmulas 2x is 2x = 2 cos2x - 1. Adding 1 on both sides, we get 1 + cos 2x = 2 cos2x. When dividing on both sides by 2, we obtain Cos2x = (1 + Cos 2x) / 2. We use this to find ?Cos2x DX. Then we stayed ? ?Cos2x DX = ? ? (1 + COS 2x) / 2 dx = (1/2) ? ? (1 + cos 2x) DX = (1/2) ? 1 DX + (1/2 ) 2x DX in the previous sections, see that ?COS 2x DX = (2x sin) / 2 + C. thus? Cos2x DX = (1/2) x + (1/2) x + / 2) x + (1/2) (Sin 2x) / 2 + C (OR) ?Cos2x DX = x / 2 + (SIN 2x) / 4 + C This is the integral of the F?oMula Cos ^ 2. Come on To prove the same formula on another method. METHOD 2: COS 2x integration using the integration by parts we know we can write cos2x as cos x ? ? Cosx. As a product, we can use the integration by parties to find the ? ?COS x ? ? COS X DX. Then we stayed ? ?Cos2x dx =? Cos x ? ? Cos x dx = ? ?U DV here, u = Cos x and DV = COS x dx. So du = - sin x dx and v = sin x. By integration by pieces of pieces, ? 'u dv = UV-' v du ? ?COS x ? ? COS X DX = (COS X) (Sin X) -? Sin X (-SIN x) DX ? ?Cos2x dx = (1/2) (2) (2 Sin x Cosx) + ? SIN2X DX by the dual angle of sin, 2 SIN X COS X = 2x sin and a trigonometic identity , SIN2X = 1 - COS2X. Then ? ?Cos2x DX = (1/2) Sin 2x + ? (1 - COS2X) DX ? ?Cos2x DX = (1/2) Sin + ? ?1 DX -? Cos2x DX ? ?Cos2x DX +? Cos2x DX = (1/2) Sin Sin + X + CA 2 ~ ? ?Cos2x DX = (1/2) Sin 2x + X + C ~? Cos2x DX = (1/4) Sin 2x + x / 2 + C (wherein C = C ^ / 2) So it proved. Defined integral of COS ^ 2x To calculate COS2x defined integral, it only replaces the upper and lower limits of the integral value and subtract the resulting values. Let's calculate some defined integral COS2X DX in here. Integral of Cos ^ 2x from 0 to 2pi ? ? ? \ (?l ^ {2 \ pi} \) Cos2x dx = [x / 2 + (SIN 2X) / 4] \ (\ left \ right | _0 ^ {2 \ pi} \) = [2I / 2 + (4I SIN) / 4] - [0 + (SEN 0) / 4] = I + 0/4 = I therefore, the COS2x integral of 0 to 2I ? I. Integral of Cos ^ 2x of 0 a PI ~ \ \ (? \ \ (^ {\ pi} \) Cos2x dx = [x / 2 + (Sin 2x) / 4] \ (\ Left \ RIGHT |. _0 ^ \ pi} \) = [I / 2 + (SIN 2I) / 4] - [0 + (sen 0) / 4] = i / 2 + 0/4 = I / 2 therefore, the COS2X integral of 0 ai It is IM / 2. Important notes related to COS 2x and integral of COS2X: COS2X COS: 2x dx = (SIN 2x) / 2 + C ?Cos2x DX = x / 2 + (SIN 2x) / 4 + C COS2X Integral and Cos2X Integral 2x: Example 1: Evaluate the Integral TM 2x Esin 2x DX. SOLUTION: Let's solve this using the replacement method. Let's sin 2x = u. Then 2 Cos 2x dx = du (or) 2x dx = (1/2) du. Then the integral given becomes a "I (1/2) du = (1/2) I + C. let us replace the value of u = sen 2x back here. So we have a ? ?Cos 2x Esin 2x DX = (1/2) Esin 2x + C. Answer: The member of COS 2x E ^ Sin 2x DX is (1/2) ESIN 2x + C. Example 2 : Evaluate the member of the 2x sin 2x dx sin. SOLUTION: by the dual fans of sin, sin 2 a co-sine A = Sin 2a. Using this, ~ ? ?COS 2x sin 2x dx = (1/2) ~ ?? 2 Sin 2x 2x dx = (1/2) ? ? ?Sin 2 (2x) dx = (1/2) ? ?Sin 4x DX Let 4x = u. Then 4 dx = du (or) dx = du / 4. then one \ ?cos 2x sin 2x dx = (1/2) ? ? u sin (du / 4) = (1/8) (1/8) (- Cos U) + C = (1/8) (- Cos 2x) + C Response: The integral of COS 2x integral sin 2x dx is (-1 / 8) (Cos 2x) + C Example 3: Evaluate the integral ? ?(1 + COS X) DX 2. SOLUTION: ? ? (1 + COS X) DX = to 2? ?(1 + 2 + Cos x Cos2x) DX = A ? \ '1 DX + 2 ~ COS DX X + A ?COS2DX We know that one ? \' Cos DX X = Sen X and A 'Cos2x DX = x / 2 + 2 x Sin) / 4. So the integral above becomes = x + 2 sen x + x / 2 + (SIN 2x) / 4 = (3x) / 2 + (1/4) Sin 2x + 2 Sen X + C. Answer: The integral of (1 + Cos X) ^ 2 is (3x) / 2 + (1/4) Sin 2x + 2 Sen X + C. View Answer> Go to Slidego to Slidego Sliding questions About Basic Mathematical Concepts? Become a troubleshooting champion using logic, not rules. Find out why these are mathematics with our certified specialists Book a class Free Trial FAQs in Cos 2x and COS ^ 2x The Cos 2x DX member is written as a ? ?Cos 2x DX and ? TM 2x dx = (Sin 2x) / 2 + C, where C is the integration constant. What is the integral of Cos ^ 2x DX? The integral of COS ^ 2 X DX is written as a ?Cos2x DX and ? ? Cos2DX = x / 2 + (SIN 2x) / 4 + C. Here, C is the integration constant . How to find the defined integral of cos 2x from 0 to PI? We know that one ? ?Cos 2x DX = (Sin 2x) / 2. Replacing limits 0 and EU, we have (1/2) Sin 2 (2I) - (1/2) Sin 2 (0) = (1/2 ) 0 - 0 = 0. What is the integral of COS ^ 3 x dx? ? ? ? ?Cos3x dx = a? COS2X COS DX X = A ? \ (1 - SIN2X) COS DX X. Let's replace SIN X = U. Then Cos DX X = du. Then the integral above becomes, ? ? (1 - U2) du = U - U3 / 3 + C. Replacing U = Sen X here behind, ?Cos3x dx = Sen X - SIN3x / 3 + C. What is the integral of Cos 3x DX? To find the ? ? ?Cos 3x DX, we assume that 3x = u. Then 3 dx = du (or) dx = du / 3. then the integral above becomes, ? ?u cos (1/3) du = (1/3) sin u + c = ( 1/3) SIN (3x) + C. How to find the defined integral of COS ^ 2x from 0 to 2pi? We know that one ? ?Cos2DX = x / 2 + (SIN 2x) / 4 + C. Replacing the limits here, we have [I / 2 + (SIN 2I) / 4] - [0 + (SIN 0) / 4] = I / 2. It is the integral cos 2x dx even if integral cos ^ DX? No, the values of these two integral are not equal. We have a ?Cos2dx = x / 2 + (SIN 2x) / 4 + C ~? COS 2x DX = (SIN 2x) / 2 + C not simply integrate cos ^ 2 (x) as it is, so we want to transform it into another form, We can easily do using trigonomic identities. Remember the Double Formula angle: COS (2x) = COS ^ 2 (x) - SIN ^ 2 (X). We also know the Trig Sin ^ 2 (x) + COS ^ 2 (x) = 1, so to combine these we obtain the COS equations (2x) = 2nd ^ 2 (x) -1. Now we can reorganize this to give: Cos ^ 2 (x) = (1 + cos (2x)) / 2. Therefore, we have a equation that gives cosm ^ 2 (x) in a more pleasant way which can easily integrate using the reverse chain rule. This will eventually give us a response from x / 2 + sen (2x) / 4 + C :) Evaluate the following: `Integral Cos int ^ 2x.dx`recall the COS 2x identity = 2 cos2x a 1, what it gives `s 2x ^ = (1 + cos2x) / (2) `So,` Cos int ^ 2 x.dx` = `(1) / (2) int (1 + cos 2x) .dx` =` (1) / (2) dx int + (1) / (2) int cos 2x .dx '= `x / (2) + (1) / (4) Sin 2x + C`. Concept: METHOD OF INTEGRATION: Integration by Substitution, Is there an error in this issue or sob? Page 2EValuate the following integral: .. `int (2) / (sqrt (x) - sqrt (x + 3)) dx``int (2) / (sqrt (x) - sqrt (x + 3)) DX` = `int (2) / (sqrt (x) - sqrt (x + 3)) xx (sqrt (x) + sqrt (x + 3)) / (sqrt (x) + sqrt (x + 3)) DX. `` = int (2 (sqrt (x) + sqrt (x + 3))) / (x - (x + 3)) dx`` = -. (2) / (3) int (sqrt (x) +. Sqrt (x + 3)) dx` `= - (2) / (3) INT X ^ (02/01) DX - (2) / (3) ) int (x + 3) ^ (02/01). DX` `= - (2) / (3) (X ^ (02/03)) / ((3/2)) - .. (2) / (3) ((x + 3) ^ (3/2 )) / ((3/2) + = `C` - (4) / (9) [x ^ (3/2) + (x + 3) ^ (02/03)] + C Concept: MET all. From the integration: integration by substitution, there is an error in this question or solution 3evaluate the following integral: `int (3) / (sqrt (sqrt 7x - 2) - SQRT (7x - 5)) DX``In (. 3) / (SQRT (7x - 2) - SQRT (7x - 5).) DX` `= int (3) / (SQRT (7x - 2) - SQRT (7x - 5)) XXA (SQRT (7x - 2) + SQRT (7x - 5)) / (SQRT (7x - 2) + SQRT (7x - 5)) DX`` = int (3 (SQRT (7x - 2) + SQRT (7x - 5))) / ((7x - 2) - (7x - 5)) DX`` = int (SQRT (7x - 2) + sqrt (7x - 5) ) .. dx` = int (7x - 2) ^ (1/2) .dx + int (7x - 5) ^ (02/01) .dx` = `((7x - 2) ^ (02/03 )) / (3/2) XX (1) / (7) + ((7x - 5) ^ (02/03)) / (3/2) XX (1) / (7) + = `C` 2) / (21) (7x - 2) ^ (3/2) + (2) / (21) (7x - 5) ^ (3/2) + c`. Concept: Method of integration o: Integration by substitution, is there an error in this matter or solution? Solution?

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