# Is it possible to integrate sin^75 x dx , or any high power of sin , like 45 , 76 , 101 etc

by Mukss 26.10.2015

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by Mukss 26.10.2015

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For odd powers of sine x we do as:

then substitute for y = cos x

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For even powers of sine x like 20 for example.. The integrand can be simplified successively. In the following let us ignore the constants and simple terms. Look at how that can be solved.

f(x) = sin²° x = (1 - cos2x)¹° / 2¹°

= 1 - 10 cos2x + 10C2 cos² 2x - 10C3 Cos³ 2x + 10C4 cos⁴ 2x - 10C5 Cos⁵ 2x +...

cos²2x = (1+cos4x)/2

cos³ 2x = (1 - sin² 2x) cos 2x , here let y = sin 2x and dy = 2 cos2x dx

cos⁴ 2x = (1+cos 4x)² / 2² = [ 1 + 2 cos 4x + (1+cos8x)/2 ] /4= 3/8+1/2*cos4x+1/8*cos8x

cos⁵ 2x = (1 - sin² 2x)² cos 2x

So now the integral will be:

I = x - 5 sin2x +10C2 * (x/2 +1/8*sin4x) - 10C3 * (y/2 - y^3/6)

+ 3x/8 +1/8*sin4x+1/64* sin 8x + ....

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Another way:

f(x) = sine²° x = (1 - cos2x )¹° / 2¹°

= 1/2¹° * (1 -2 cos 2x + cos² 2x)⁵

= 1/2¹° * [1 - 2 cos 2x + (1+cos4x)/2 ]⁵

= 1/2¹° * [ (3/2 - 2 cos 2x + 1/2 cos 4x)² ]² (3/2 - 2 cos 2x + 1/2 cos 4x)

= 1/2¹° * [ 9/4 + 4cos² 2x + 1/4 * cos²4x - 6cos2x -2 cos2x cos4x +3/2 cos4x ]² *

* (3/2 - 2 cos2x + 1/2 cos 4x)

= 1/2¹° * [ 9/4 + 2(1+cos4x) +1/8 (1+cos8x) - 3 (1+cos4x) - (cos6x+cos2x) +3/2 cos4x]² *

* (3/2 - 2 cos2x + 1/2 cos 4x)

= 1/2¹° * [ 11/8 - cos 2x + 1/2 cos4x - cos 6 x +1/8 * cos 8x ]² * (3/2 -2cos2x +1/2cos4x)

Multiplications can be done and then simple terms can be integrated. trigonometric formulae to be used for cosine cosine products..

then substitute for y = cos x

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==========================================

For even powers of sine x like 20 for example.. The integrand can be simplified successively. In the following let us ignore the constants and simple terms. Look at how that can be solved.

f(x) = sin²° x = (1 - cos2x)¹° / 2¹°

= 1 - 10 cos2x + 10C2 cos² 2x - 10C3 Cos³ 2x + 10C4 cos⁴ 2x - 10C5 Cos⁵ 2x +...

cos²2x = (1+cos4x)/2

cos³ 2x = (1 - sin² 2x) cos 2x , here let y = sin 2x and dy = 2 cos2x dx

cos⁴ 2x = (1+cos 4x)² / 2² = [ 1 + 2 cos 4x + (1+cos8x)/2 ] /4= 3/8+1/2*cos4x+1/8*cos8x

cos⁵ 2x = (1 - sin² 2x)² cos 2x

So now the integral will be:

I = x - 5 sin2x +10C2 * (x/2 +1/8*sin4x) - 10C3 * (y/2 - y^3/6)

+ 3x/8 +1/8*sin4x+1/64* sin 8x + ....

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Another way:

f(x) = sine²° x = (1 - cos2x )¹° / 2¹°

= 1/2¹° * (1 -2 cos 2x + cos² 2x)⁵

= 1/2¹° * [1 - 2 cos 2x + (1+cos4x)/2 ]⁵

= 1/2¹° * [ (3/2 - 2 cos 2x + 1/2 cos 4x)² ]² (3/2 - 2 cos 2x + 1/2 cos 4x)

= 1/2¹° * [ 9/4 + 4cos² 2x + 1/4 * cos²4x - 6cos2x -2 cos2x cos4x +3/2 cos4x ]² *

* (3/2 - 2 cos2x + 1/2 cos 4x)

= 1/2¹° * [ 9/4 + 2(1+cos4x) +1/8 (1+cos8x) - 3 (1+cos4x) - (cos6x+cos2x) +3/2 cos4x]² *

* (3/2 - 2 cos2x + 1/2 cos 4x)

= 1/2¹° * [ 11/8 - cos 2x + 1/2 cos4x - cos 6 x +1/8 * cos 8x ]² * (3/2 -2cos2x +1/2cos4x)

Multiplications can be done and then simple terms can be integrated. trigonometric formulae to be used for cosine cosine products..