Integral of sin^5(2x)cos^2(2x)dx

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Integral of sin^5(2x)cos^2(2x)dx

Powers of Sin and Cos Example: Evaluate int sin5x dx we write Exercise: int cos7x dx Example: Evaluate int sin4x dx Solution: We write int sin4x dx = int (sin2x)2 dx = int (1/2 - 1/2 cos(2x))2 dx = int 1/4 - 1/2 cos(2x) + 1/4 cos2(2x) dx For the second integral let u = 2x dx = 1/2du = 1/4 x - 1/4 sin(2x) + 1/4 int 1/2 + 1/2 cos(4x) Let u = 4x dx = 1/4du = 1/4 x

- 1/4 sin(2x) + 1/8 x + 1/32 sin(4x) + C We see that if the power is odd we can pull out one of the sin functions and convert the other to an expression involving the cos function only. Then use u = cos x. If the power is even, we must use the trig identities sin2x = 1/2 - 1/2 cos(2x) and cos2x = 1/2 + 1/2 cos(2x) This method will always work and is always long

and tedious. Mixed powers of Sin and Cos Example: Evaluate int sin2x cos3x dx. We pull out a cos x and convert the cos2 x to 1 - sin2 x: int sin2x (1 - sin2 x) cosx dx Let u = cosx, du = -sinx dx We have int u2(1 - u2)du = int u2 - u4 du = 1/3 u3 - 1/5 u5 + C = 1/3 cos3x - 1/5 cos5x + C

Exercise: int sin5x cos2x dx Powers of Tangents and Secants

To integrate powers of tangents and secants we use the formula tan2 x = sec2 x - 1 Example: Evaluate int tan4 x dx Solution: We write int tan2 x tan2 x dx = int tan2 x (sec2 x - 1) = int tan2 x sec2 x dx - int tan2 x dx = int tan2 x sec2 x dx - int (sec2 x - 1) dx For the first integral let u = tan x du = sec2 x dx. We have int u2 du - tanx - x = 1/3 u3 - tanx - x + C

= 1/3 tan3 x - tan x - x + C Exercise: int sec5 x tan3 x dx. Mixed Angles

We have the following formulas: sin(mx) sin(nx) = 1/2[cos[(m - n)x] - cos[(m + n)x]] sin(mx) cos(nx) = 1/2[sin[(m - n)x] + sin[(m + n)x]] cos(mx) cos(nx) = 1/2[cos[(m - n)x] + cos[(m + n)x]] Example: int sin(3x)sin(4x)dx = 1/2 int cos(-x) - cos(7x) dx We integrate the first by letting u = -x

and the second by letting u = 7x. We found a book related to your question. SEE SOLUTIONS We found a book related to your question. SEE SOLUTIONS We found a book related to your question. SEE SOLUTIONS C?u h?i n��y thu?c ?? thi tr?c nghi?m d??i ??y, b?m v��o B?t ??u thi ?? l��m to��n b��i YOMEDIA H??ng d?n Tr?c nghi?m Online v�� T��ch l?y

?i?m th??ng C?U H?I KH?C To integrate this we need to use the chain rule, substituting cos2x = u Integral becomes: u2sin32x dxChain rule: dy/dx = du/dx dy/du du/dx = -2sin2x --> dx = -1/2sin2x du Substituting into the equation u2sin32x * -1/2sin2x du Simplifies to: -2u2sin22x duWe know that cos2x + sin2x =1 Integral = -2u2(1 - cos22x) du Substituting ->

-2u2 + 2u4 duIntegrating this: -2( 1/3u3 - 1/5u5) + c Substituting u back into the equation: cos52x/10 - cos32x/6 + c If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains *. and *. are unblocked.

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