Question

integral sinx upon 1+ cosx^2

Answer 1

this is the answer ...

Answer 2

Hey there!

Answer:



$I = \int \dfrac{sinx}{(1 + cosx)^{2}} dx\\ \\ Let\: this\: be\: the\: first\: equation \\ \\ Put \: 1 + cosx = t \\ \\ Now\: differentiate \:both\: sides\: wrt x\\ \\ - \: sinx \:dx = dt \\ \\ sinx \:dx = - dt\\ \\ Putting\: in \: equation \: ( 1 ) \\ \\I = \int \dfrac{- dt}{t^{2}} \\ \\I = - \int \dfrac{1}{t^{2}} dt\\ \\ I = - \int t^{-2} dt\\ \\ I = - \dfrac{t^{-1}}{-1} \\ \\ I = t^{-1} \\ \\ I = \dfrac{1}{t} \\ \\ Substituting\: value \: of \: t \\ \\ I = \dfrac{1}{1 + cosx} + C$



Where C is the Arbitrary constant.



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