Evaluating ∫cot⁡xcsc2xdx with u=cot⁡x Differentiating both sides of u, then making the substitution: u=cot⁡x du=−cot⁡xcsc⁡xdx, dx=−duucsc⁡x ∫−ucsc2xduucsc⁡x=∫−csc⁡xdu Apparently, this was not an adequate approach, because x is still part of the integrand. What should be done instead?

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Evaluating ∫cot⁡xcsc2xdx with u=cot⁡x  Differentiating both sides of u, then making the substitution:  u=cot⁡x   du=−cot⁡xcsc⁡xdx,    dx=−duucsc⁡x   ∫−ucsc2xduucsc⁡x=∫−csc⁡xdu  Apparently, this was not an adequate approach, because x is still part of the integrand. What should be done instead?

ataill0k · Accepted Answer

You have du=−csc2x dx rather than your wrong differentiation. This implies the integral is  ∫cot⁡xcsc2xdx=∫−udu=−12u2+c=−12cot2x+c  On the other hand, rewriting the integral as  ∫cos⁡xsin3x dx=∫(sin⁡x)−3d(sin⁡x)=−121sin2x+c  is much easier.

dikgetse3u · Accepted Answer

For alternative way:  ∫cot⁡xcsc2xdx  =∫cos⁡xdxsin3x  Now you can advance taking sin⁡x=z

Evaluating \int \cot x \csc^2x dx with u=\cot x Differentiating both sides of u, the

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