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

1
by Mukss

2015-10-27T20:03:39+05:30

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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..
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