How do I integrate cos7x-cos8x/1+2cos5x?
First of all solve the equation( cos 7x - cos 8x ) = ( cos 7x + cos 3x ) - ( cos 8x + cos 2x ) + ( cos 2x - cos 3x ) = ( 2 cos 5x cos 2x ) - ( 2 cos 5x cos 3x ) + ( cos 2x - cos 3x ) = ( 2 cos 5x )( cos 2x - cos 3x ) + ( cos 2x - cos 3x ) = ( cos 2x - cos 3x ) ( 2 cos 5x + 1 )divided it by 1+2cos5xso we get= cos2x-cos3x.now integrate that and we get(1/2)sin2x - (1/2)sin3x + C.This is the answer.
How do I integrate [cos(7x)-cos(8x)] / [1+2cos(5x)]?
Use multiple angle formulas and you will find the integrand can be expressed in terms of cox(x)^n, and can then be integrated to give-sin(3*x)/3+sin(2*x)/2, which can be confirmed by differentiation and multiple angle formulas.
How do I integrate 1/(cos^(2n))?
let 2n =m and 1/cos ^2n[x] =sec ^m [x] now integrate sec ^m [x] dxsplit sec ^m[x] = sec^[m-2]x .sec ^2 x dx integrate by partslet u= sec^[m-2] and du = sec^2x dxthen I = sec ^[m-2] . tan x - integral of v du , note du = [m-2] sec^[m-3] x. sec x tan x[ using chain rule , and remember d/dx [ sec x ] = sec x . tan x ]I[1] = sec ^[m-2]x.tanx- int of tan x.[m-2]sec ^ [m-2] x . tan^ x , eq 1at this point just concider the 2 nd part of I[1]I[2] =- [m-2] int of sec^[m-2] x .tan^ 2 x d x since tan ^2 x = sec ^2 x -1I[2] =-[m-2] int of sec ^[m-2] x . [sec ^ 2 x -1] d x , distribute inside the bracketsI[2] =- [m-2] int of sec ^ m [x] +[m-2] int of sec ^ [m-2] x dxnow bring -[m-2] int of sec ^m x to the lhs of I[1] and eq 1 and add it to the int of sec ^ m x d xto get [m-1] int of sec ^ m x dx = tan x. sec ^[m-2] x + [m-2] int of sec^[m-2] x d xdivide across by [m-1] to getI[1] = int of sec ^m x dx = (1/[m-1]) [tan x ][sec ^[m-2] x] + ([m-2]/[m-1]) int of sec^[m-2] x dxIt looks somewhat random but I am sure when it is written out and m=2n it will come together nicely.
How do you integrate 4sin²4x?
whenever you are integrating a sine or cosine squared, rewrite it with one of the double angle identities for cosine. From cos 2x = 1 - 2sin²x if you solve for sin²x, you get: sin²x = (1-cos2x) / 2 sin²x = 1/2 - 1/2cos2x So for your expression, you can multiply by 2 and substitute 4x for x to yield: 4sin²(4x) = 2 - 2cos8x Integrate: ∫4sin²(4x) dx =∫2 - 2cos8x dx = 2x - 1/4sin8x + C
How do I integrate sinx cosx cos2x cos4x cos8x dx?
We have,[math] \displaystyle\int (\underbrace{\sin x \cos x} \cos 2x \cos 4x \cos 8x) dx [/math][math]=\frac{1}{2}× \displaystyle\int (\underbrace{\sin 2x \cos 2x} \cos 4x \cos 8x ) dx [/math][math] [\text{As, }2\sin \theta \cos \theta = \sin 2\theta, \forall \theta \in \mathbb{R}.][/math][math]=\frac{1}{2^2} ×\displaystyle\int (\underbrace{\sin 4x \cos 4x} \cos 8x ) dx [/math][math]=\frac{1}{2^3} ×\displaystyle\int (\underbrace{\sin 8x \cos 8x}) dx [/math][math]=\frac{1}{2^4}×\displaystyle\int \sin 16x \quad dx [/math][math]=\frac{1}{16} × \left(-\frac{\cos 16x}{16} \right) +c ; \text{ where, c is constant of integration.} [/math][math] \boxed{=-\frac{\cos 16x}{256}+c.}\quad \dagger [/math]Hope, you'll understand..!!P.S.- Feel free to comment if you have any query with it.Thank You!
What is the integration of sin^4x.cos^4x?
Hi,Please refer below snapshotAll the bestKKG
Can anyone help me Integrate cos^4 (4x) dx?
The (4x) is throwing me off a bit. I could figure it out if it's just (x), but...I'm getting confused. Here is what I have so far: ∫cos^4 (4x) dx = ∫ (cos²4x)² dx = ∫ [ ( 1+cos(8x) ) / 2 ]² dx = (1/4) ∫ (1+cos(8x) )² dx = (1/4) ∫ [1 + 2cos(8x) + cos²(8x) ] dx = (1/4) [ ∫ dx + ∫ 2cos(8x) dx + ∫ cos²(8x) dx ] = (1/4) [ ∫ dx + ∫ 2cos(8x) dx + (1/2) ∫ 1 + cos(16x) dx ] = (1/4) [ x + (1/4) sin(8x) + (1/2) ( ∫ dx + ∫ cos(16x) dx) ] = (1/4) [ x + (1/4) sin(8x) + (1/2) ( x + (1/16) sin(16x) ) ] = (1/4)x + (1/16) sin(8x) + (1/8)x + (1/128) sin(16x) My Final Answer: = (3/8)x + (1/16) sin(8x) + (1/128) sin(16x) + C I'm so confused. My professor gave us the correct answer: (3/8)x + (1/16) sin(8x) + (1/144) sin(16x) + C. I'm only freakin' one term off, and I can't figure out where I made my mistake. Can anyone help me?
What is the integral of sin^2(4x)??? thanks?
∫sin^2 (4x) dx = (1/2) ∫2sin^2 (4x) dx = (1/2) ∫( 1- cos (8x) ) dx = (1/2) [x - (1/8)sin(8x)] + c.