Use substitution to evaluate the integral: the integral of Π/4 to 3Π/4 of cot x dx So far, I have that u = cot x , du = -csc x squared, and dx = -1/(cscx squared) Thanks for any help
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Originally Posted by turtle Use substitution to evaluate the integral: the integral of Π/4 to 3Π/4 of cot x dx So far, I have that u = cot x , du = -csc x squared, and dx = -1/(cscx squared) Thanks for any help The integral, $\displaystyle \int \cot x dx = \int \frac{\cos x}{\sin x}dx$ Let, $\displaystyle u=\sin x$. Thus, $\displaystyle -\int \frac{1}{u} du$ Thus, $\displaystyle -\ln |u|+C=-\ln |\sin x|$
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