(integral sign) 2/(x^2-2x-8)dx (integral sign) sin^3(x)cos^(3/2)(x)dx Solution U

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Use; sin 2x = 2 sin x * cos x & cos 3x = (4 cos^3 x - 3 cos x) so, sin^3 (2x) cos^2 (3x) = (2 sin x * cos x)^3 * (4 cos^3 x - 3 cos x)² => 8 sin^3 x * cos^3 x * (16 cos^6 x + 9 cos² x - 24 cos^4 x ) => 8 sin x (1 - cos² x) * cos^3 x * (16 cos^6 x + 9 cos² x - 24 cos^4 x ) now let I = ?sin^3 (2x) cos^2 (3x) dx => ? 8 sin x (1 - cos² x) * cos^3 x * (16 cos^6 x + 9 cos² x - 24 cos^4 x ) dx putting cos x = t therefore , sin x dx = - dt => I = - 8 ? (1 - t²) * t^3 (16 t^6 + 9 t² - 24 t^4) dt => I = - 8 ? (1 - t²) * (16 t^9 + 9 t^5 - 24 t^7) dt => I = - 8 ? (16 t^9 + 9 t^5 - 24 t^7 - 16 t^11 - 9 t^7 + 24 t^9 ) dt => I = - 8 ? (40 t^9 + 9 t^5 - 33 t^7 - 16 t^11 ) dt => I = 8 ? (16 t^11 - 40 t^9 + 33 t^7 - 9 t^5 ) dt => I = 8 [ 16 * (t^12)/12 - 40 * (t^10)/10 + 33 * (t^8)/8 - 9 * (t^6)/6 + C => I = (32/3) *(t^12) - 32 * (t^10) + 33 * (t^8) - 12 * (t^6) + C NOW resubstituting t = cos x => I = (32/3) *(cos^12 x ) - 32 * (cos^10 x ) + 33 * (cos^8 x) - 12 * (cos^6 x ) + C ------------> (Ans) ==================================== Alternatively=> it can be solved like this also. sin^3 (2x) cos^2 (3x) = sin 2x * (sin 2x * cos 3x )² => sin 2x * {(sin 5x - sin x )/2 }² => (1/4) sin 2x * (sin² 5x + sin² x - 2 sin 5x * sin x) => (1/4) sin 2x * [(1 - cos 10x)/2 + (1 - cos 2x)/2 - 2 sin 5x * sin x) => (1/4) sin 2x * [(1 - cos 10x)/2 + (1 - cos 2x)/2 - (cos 4x - cos 6x) which on further simplification it can be splitted in single terms. Note : I checked it out and found no mistakes. only the difference is , answer from Wolfram is comming in multiple angles form. here is the link. http://integrals.wolfram.com/index.jsp?e…

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