Is it possible to integrate sin^75 x dx , or any high power of sin , like 45 , 76 , 101 etc
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For odd powers of sine x we do as:
then substitute for y = cos x
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[tex] I= \int\limits^{}_{} {sin^{45}x} \, dx = \int\limits^{}_{} {(1-cos^2x)^{22}\ sinx} \, dx\\\\let\ y =cosx,\ \ dy=-sinx\ dx\\\\I= -\int\limits^{}_{} {(1-y^2)^{22}} \, dy \\\\.= -\int\limits^{}_{} {(1-22y^2+{}^{22}C_2*y^4-{}^{22}C_3*y^6+...+(y^2)^{22})} \, dy \\\\.I=-y+22*y^3/3-{}^{22}C_2*y^5/5+...[/tex]
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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
========================================
[tex] I= \int\limits^{}_{} {sin^{45}x} \, dx = \int\limits^{}_{} {(1-cos^2x)^{22}\ sinx} \, dx\\\\let\ y =cosx,\ \ dy=-sinx\ dx\\\\I= -\int\limits^{}_{} {(1-y^2)^{22}} \, dy \\\\.= -\int\limits^{}_{} {(1-22y^2+{}^{22}C_2*y^4-{}^{22}C_3*y^6+...+(y^2)^{22})} \, dy \\\\.I=-y+22*y^3/3-{}^{22}C_2*y^5/5+...[/tex]
==========================================
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 + ....
================================
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..
kvnmurty:
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