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Shinoy Philip Grade: 12
        

sin (tan-1 x)/1 + x2


please solve the above question


 

6 years ago

Answers : (6)

AKASH GOYAL AskiitiansExpert-IITD
419 Points
										

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


AskiitiansExpert-IITD


 


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6 years ago
aditya suman
39 Points
										

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

6 years ago
vikas askiitian expert
510 Points
										

 I=(sintan-1x) /1+x2


put tan-1x =t


   differentiating wrt x


1/1+x2 dx =dt


I=sintdt=-cost + c


  =-costan-1x + c

6 years ago
arvind ramesh
32 Points
										

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

6 years ago
rajan jha
49 Points
										

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?

6 years ago
Souradeep Majumder
80 Points
										

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..................

6 years ago
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