Evaluate Definite Integral = _______
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regular substitution u=x^2....du=2x dx integral(.5cos(u)) should be easy
Originally Posted by tbenne3 Evaluate Definite Integral = _______ use the method of substitution ... let $\displaystyle u = x^2$ $\displaystyle du = 2x \, dx$ $\displaystyle \frac{1}{2} \int_0^{\sqrt{\pi}} 2x \cos(x^2) \, dx$ substitute, reset the limits of integration, integrate and evaluate ...
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