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Find the integral of (x² +cos x)dx |
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Answer» The answer is = (x2 - 2) sin x+2x cos x + C Explanation: The integration by parts is ∫uv'dx = uv - ∫u'v Apply the integration by parts Let u = x2, ⇒ , u' = 2x v' = cos x, ⇒ , v = sin x Therefore, ∫x2 cos x dx = x2 sin x - ∫2x sin x dx Apply the integration by parts a second time Let u = x, ⇒ , u' = 1 v' = sin x, ⇒ , v = -cos x So, ∫x2 cos x dx = x2 sin x - ∫2x sin x dx = x2 sin x - 2(-x cos x - ∫-cos x dx) = x2 sin x + 2x cos x - 2 sin x + C = (x2 - 2) sin x + 2x cos x + C |
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