InterviewSolution
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InterviewSolution
| 1. |
integrate sin^4x.dx |
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Answer» ∫sin^4(x) => ∫ [sin^2(x)]^2 => ∫ [(1 -cos(2x)/2)]^2 =>1/4 ∫[1 + cos^2(2x) - 2cos(2x)] =>1/4 [1 - 2cos(2x) + (1 + cos(4x)/2 ] => ∫ (1/4) - ∫(1/2)cos(2x) + ∫(1/8) + ∫(1/8)cos(4x) => ∫(3/8) - ∫(1/2)cos(2x) +∫ (1/8)cos(4x) so ∫sin^4(x)dx = 3/8∫dx - 1/2∫cos(2x) dx + 1/8∫cos(4x) dx ∫sin^4(x)dx = (3/8)x - (1/2)sin(2x)*(1/2) + (1/8)sin(4x)*(1/4) + c ∫sin^4(x)dx = (3/8)x - (1/4)sin(2x) + (1/32)sin(4x) + c |
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