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Evaluate the definite integral int on [pi,0] x cos(x^2)dx?

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  • Evaluate the definite integral int on [pi,0] x cos(x^2)dx?


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lnt[x cos(x^2)]dx Let u = x^2 Then du = 2xdx dx = du/2x = lnt[ xcos(u)*du/2x] = lnt[cos(u)du/2] (pi)^2 to 0 = sin(u)/2 pi^2 ...
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Positive: 86 %
... of definite integrals that \displaystyle{\int_a^b f(x) dx ... dfrac{\sin x}{\cos x} \right )} dx ... definite integral [math]\displaystyle \int_0^ ...
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Positive: 83 %

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Positive: 44 %

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