prove that integration of (sin x)^3 is= -3/4(cos x) + 1/12(cos 3x) + c?

Question

using trigo identities and reduction formula for (sin x)^3

Answer 1

I = ∫ sin ³ x dx
I = ∫ (sin x) (sin ² x) dx
I = ∫ (sin x) (1 - cos ² x) dx
Let u = cos x
du = - sin x dx
- du = sin x dx
I = - ∫ 1 - u ² du
I = - ( u - u ³ / 3 ) + C
I = - cos x + cos ³ x / 3 + C

Answer 2

Since the derivative of the right side is
= d/dx [-3/4 (cos x) + 1/12 (cos 3x) + C]

= -3/4 (-sin x) + 1/12(-3 sin 3x)

= -3/4 (-sin x) - 1/4 (sin 3x)

= 3/4 sin x - 1/4 ( 3 sin x - 4 sin^3 x)

= 3/4 sin x - 3/4 sin x + sin^3 x

= sin^3 x

= (sin x)^3

and is equal to the right side, we conclude that
integral of (sin x)^3 through x = -3/4(cos x) + 1/12(cos 3x) + c

:D

Answer 3

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RE:
prove that integration of (sin x)^3 is= -3/4(cos x) + 1/12(cos 3x) + c?
using trigo identities and reduction formula for (sin x)^3

Answer 4

Integral Of Sin 3

Answer 5

Hey friend,
NOTE:
* sin 3x = 3sinx - 4 (sin x)^3
therefore, (sin x)^3 = (3/4)sin x - (1/4)sin 3x
*Integration ( sin x ) = - cos x
*Integration ( sin 3x) = -(1/3)cos 3x
PROOF:
Integ( sin x)^3 = Integ{ (3/4)sin x } - Integ { (1/4)sin 3x} + C
..................... = (3/4)(- cos x) - { (1/4)[-1/3 (cos 3x)] } + C
..................... = -3/4(cos x) + 1/12(cos 3x) + C
Hence Proved

Answer 6

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Proving the integral is hard, but the differential is easier. Use this to prove the integration of the aforementioned function. If y = asin(x/a), then x/a = sin(y), where a is an arbitrary constant. dx/a = cos(y) dy 1/cos(y) * 1/a = dy/dx (This is the value you want!) dy/dx = 1/sqrt(1 - sin^2(y) * 1/sqrt(a^2) Note: a^2*sin^2(y) = x^2 dy/dx = 1/sqrt(a^2 - x^2) The value you put in there is the differential of the hyperbolic arcsine of x.

Answer 7

∫sin³(x)dx
= ∫sin²(x)sin(x)dx
= ∫(1 - cos²(x))sin(x)dx
= ∫sin(x)dx - ∫cos²(x)sin(x)dx

∫sin(x)dx
= -cos(x)

∫cos²(x)sin(x)dx
Let u = cos(x)
du = -sin(x)dx

∫cos²(x)sin(x)dx
= ∫-u²du
= -u³/3 + c
= -(1/3)cos³(x) + c

∫sin(x)dx - ∫cos²(x)sin(x)dx
= -cos(x) + (1/3)cos³(x) + c

cos(2x) = cos²(x) - sin²(x)
sin(2x) = 2sin(x)cos(x)
cos(3x) = cos(2x+x)
= cos(2x)cos(x) - sin(2x)sin(x)
= (cos²(x) - sin²(x))cos(x) - 2sin²(x)cos(x)
= cos³(x) - sin²(x)cos(x) - 2sin²(x)cos(x)
= cos³(x) - 3sin²(x)cos(x)
= cos³(x) - 3(1-cos²(x))cos(x)
= cos³(x) - 3cos(x) + 3cos³(x)
= 4cos³(x) - 3cos(x)

4cos³(x) = cos(3x) + 3cos(x)
cos³(x) = (1/4)cos(3x) + (3/4)cos(x)

-cos(x) + (1/3)cos³(x) + c
= -cos(x) + (1/12)cos(3x) + (1/4)cos(x) + c
= -(3/4)cos(x) + (1/12)cos(3x) + c

Answer 8

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