Answer to Question #108785 in Calculus for Nimra

Question #108785

∫csc(ax)dx


1
Expert's answer
2020-04-14T18:16:30-0400

Answer: csc(ax)  dx=1alncsc(ax)  +  cot(ax)  +  C\int csc(ax)\;dx=-\frac1a\ln\left|csc(ax)\;+\;cot(ax)\right|\;+\;C\\

csc(ax)  dx=1aacsc(ax)  dx  u=axdu=adxSubstitute  u.1aacsc(ax)  dx  =  1acsc  u  du  csc  u  du  =    csc  ucsc  u  +  cot  ucsc  u  +  cot  udu==csc2  u  +  csc  ucot  ucsc  u  +  cot  uduv  =  csc  u  +  cot  uLets  take  derivative  of  v.  dv=(csc  ucot  u  +  csc2u)  duSubstitute  v.(csc2  u  +  csc  ucot  u)  ducsc  u  +  cot  u=dvv==lnv+C==lncsc  u  +  cot  u+C  ==  lncsc(ax)  +  cot(ax)+Ccsc(ax)  dx=1alncsc(ax)  +  cot(ax)  +  C\int csc(ax)\;dx=\frac1a\int a\cdot csc(ax)\;dx\;\\u=ax\Rightarrow du=adx\\Substitute\;u.\\ \frac1a\int a\cdot csc(ax)\;dx\;=\;\frac1a\int csc\;u\;du\;\\ \int csc\;u\;du\;=\;\;\int csc\;u\frac{csc\;u\;+\;cot\;u}{csc\;u\;+\;cot\;u}du=\\ =-\int-\frac{csc^2\;u\;+\;csc\;u\cdot cot\;u}{csc\;u\;+\;cot\;u}du\\ v\;=\;csc\;u\;+\;cot\;u\\ Let's\;take\;derivative\;of\;v.\;\\ dv=-(csc\;u\cdot cot\;u\;+\;csc^2u)\;du\\ Substitute\;v.\\ -\int-\frac{(csc^2\;u\;+\;csc\;u\cdot cot\;u)\;du}{csc\;u\;+\;cot\;u}=-\int\frac{dv}v=\\=-\ln\left|v\right|+C=\\ =-\ln\left|csc\;u\;+\;cot\;u\right|+C\;=\\=\;-\ln\left|csc(ax)\;+\;cot(ax)\right|+C\\ \int csc(ax)\;dx=-\frac1a\ln\left|csc(ax)\;+\;cot(ax)\right|\;+\;C\\

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