# sin (tan-1 x)/1 + x2please solve the above question

420 Points
13 years ago

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

Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.

Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

39 Points
13 years ago

put tan-1x=t

then , dt/dx=1/1+x2

dt=dx/1+x2                _1

now ,sintdx/1+xusing 1

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

509 Points
13 years ago

I=(sintan-1x) /1+x2

put tan-1x =t

differentiating wrt x

1/1+x2 dx =dt

I=sintdt=-cost + c

=-costan-1x + c

arvind ramesh
32 Points
13 years ago

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

rajan jha
49 Points
13 years ago

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).

80 Points
13 years ago

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.