sin (tan^-1 x)/1 + x²

please solve the above question

dear Shinoy

put tan^-1x = t

(1/1+x²)dx =dt

put this

integral becomes ∫sint dt = -cost + C

                                  = -cos(tan^-1x) + C   Answer

All the best.

AKASH GOYAL

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put tan^-1x=t

then , dt/dx=1/1+x²

dt=dx/1+x^{2                _}1

now ,sintdx/1+x²  using 1

=sintdt=-cost +c=-cos(tan^-1x)+c

 I=(sintan^-1x) /1+x²

put tan^-1x =t

   differentiating wrt x

1/1+x² dx =dt

I=sintdt=-cost + c

  =-costan^-1x + c

sub tan-1x as t....dt is 1/1+x^2 dx =>intg sint dt =>ans.= -cos(tan-1x)+c;

let (tan-1 x)=t

then dt/dxsqrt()=1/1 + x2

then the integration becomes

 sin(t)dt =-cos(t)

=-cos((tan-1 x))

=-cos(sec-1sqrt(1 + x2))

= -cos(cos-1sqrt(1/1+x2))

=-sqrt((1/1+x2).

hence the answer

like my answer?

put tan^-1x =z..  then sin^-1x dx =dz.  thaen it is in the form integration sinzdz and ans is -cosz+c put z=tan^-1x and you will find the value  -cos(tan^-1x)+c.

approve my answer..................