 # sin (tan-1 x)/1 + x2please solve the above question AKASH GOYAL AskiitiansExpert-IITD
420 Points
12 years ago

dear Shinoy

put tan-1x = t

(1/1+x2)dx =dt

put this

integral becomes ∫sint dt = -cost + C

All the best.

AKASH GOYAL

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12 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

12 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

12 years ago

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

12 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).