Integrate: 1.  Sin-1x / (1-x2)3/2 2. …

3) 

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

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

Ans. 1)

\(\int{ {sin ^{-1}x}dx \over ({{1-x^2})^{3\over2}}} => \int{ {sin ^{-1}x}dx \over {({1-x^2}) \sqrt {({1-x^2)}}}}\)

Put \(sin^{-1} x = t => {1\over \sqrt{({1-x^2})}}dx = dt \)

=> \(\int {t dt \over (1-sin^2t)} = \int {t dt \over cos^2t} = \int {t .sec^2tdt }\)

Using Integaration By Parts, We get 

=> t tan t + log (cos t) +  C 

=> \(sin ^{-1}x. tan (sin ^{-1}x) + log (cos (sin^{-1}x)) + C\)

2)\(\int {log \space x \space dx \over x^n}\)

Using Integration by Parts

=> \(log \space x .\int {1\over x^n}dx - \int[ { 1\over x} . \int {1\over x^n}dx]dx + C\)

=> \(log \space x .{(x^{-n+1})\over (-n+1)} - \int {1\over x}.{(x^{-n+1})\over (-n+1)} dx + C\)

=> \(log \space x .{(x^{-n+1})\over (-n+1)} - {1\over (1-n)}\int {1\over x^{n} }dx + C\)

=> \(log \space x .{(x^{-n+1})\over (-n+1)} - {1\over (1-n)^2} {1\over x^{n-1} } + C\)