The integral of the function cot(x)ln(sin(x)) can be evaluated using basic integration techniques without employing integration by parts, partial fractions, or trigonometric substitution. By substituting u = sin(x), the integral simplifies to ∫(ln(u)/u) du, which can be solved directly. The final result is (ln(sin(x)))²/2 + C, where C represents the constant of integration. This method effectively utilizes substitution and the properties of logarithms to arrive at the solution.
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Where's the "dx" ?whatlifeforme said:Homework Statement
evaluate the integral.
Homework Equations
\int (cotx)[ln(sinx)]
how would i do this using basic integration techniques from calculus 1? i am not allowed to use, int by parts, partial fractions, trig sub, etc..
The Attempt at a Solution
i tried doing a u=sinx but don't know how to get the cotx.
whatlifeforme said:thanks.
so i have:
u=sinx; du=cosx
\int \frac{ln(u)}{u} du
v=lnu; dv=(1/u) du
\int v dv = \frac{v^2}{2} = \frac{(lnu)^2}{2}
= \frac{[ln(sinx)]^2}{2}