integral from sqrt(pi/2) to sqrt(pi) of 9t^3cos(t^2)dt
I know there has to be a 2nd substitution, which is what confuses me. Please do a step by step.
first, rewrite it and divide and multiply it by 2 as:
9(1/2) ∫ t² [2t cos(t²)] dt = now let [2t cos(t²)] dt = dv; t² = u → cos(t²) (2t dt) = dv → cos(t²) d(t²) = dv → sin(t²) = v; 2x dxt = du then, integrating by parts, 9(1/2) ∫ t² [2x cos(t²)] dt= (1/2) [t²sin(t²) - ∫ sin(t²) 2t dt] = 9(1/2) t²sin(t²) - (1/2) ∫ sin(t²) 2t dx = 9(1/2) t²sin(t²) - (1/2) ∫ sin(t²) d(t²) = 9(1/2) t²sin(t²) - (1/2) [ -cos(t²)] + c = 9(1/2) t²sin(t²) + (1/2) cos(t²) + c
please reply if you want help........................................
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