Integrate: ___Sinx+1___  dx Sinx (1+ cosx)

Ans. \(\int {({sin x + 1})dx \over sinx (1+cos x)} = \int {sin x dx \over sinx (1+cos x)} + \int { dx \over sinx (1+cos x)}\)

Let \(I = I_1 + I_2\)

Where \(I= \int {({sin x + 1})dx \over sinx (1+cos x)}, I_1 = \int {sin x dx \over sinx (1+cos x)} \space and \space I_2= \int { dx \over sinx (1+cos x)}\)

=> \(I_1= \int {sin x dx \over sinx (1+cos x)} = \int { dx \over (1+cos x)} \)

=> \( \int { dx \over 2cos^2 {x \over 2}} = {1\over 2}\int {sec^2 {x \over 2}dx} \)     

=> \({2\over 2}{tan^2 {x \over 2} + C} = {tan^2 {x \over 2} + C} \space \space \space \space \space \space \space (1)\)

Now, 

 \( I_2= \int { dx \over sinx (1+cos x)} = \int { dx \over {2tan{x\over 2}\over (1+tan^2{x\over2})} ({2cos^2{x\over 2}})}\)

=> \({1\over 4} \int {({1+tan^2{x\over2})} ({sec^2{x\over 2}}) \over {tan{x\over 2}}}dx\)

\(Put \space tan {x\over2} = t \space\space\space\space => {1\over 2} sec^2{x\over 2} dx = dt \), Now 

\(=> {1\over 2} \int {({1+t^2)} \over t}dt = {1\over 2} \int {{1} \over t}dt + {1\over 2} \int {{t^2} \over t}dt \)

=> \({1\over 2} \int {{1} \over t}dt + {1\over 2} \int {{t}.}dt \)

=> \( {1\over 2} log\space t + {1\over 2} {{t^2} \over 2} = {1\over 2} log\space t + {1\over 4} {{t^2}}\)

=> \( {1\over 2} log\space (tan {x\over 2}) + {1\over 4} {tan^2{x\over 2}} \space \space \space \space \space \space \space (2)\)

From (1) And (2), We get 

I = \({tan^2 {x \over 2} } + {1\over 2} log\space (tan {x\over 2}) + {1\over 4} {tan^2{x\over 2}} + C \)

=> \(I = {1\over 2} log\space (tan {x\over 2}) + {5\over 4} {tan^2{x\over 2}} +C\)