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```        sin (tan-1 x)/1 + x2
```
7 years ago

419 Points
```										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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```
7 years ago
39 Points
```										put tan-1x=t
then , dt/dx=1/1+x2
dt=dx/1+x2                _1
now ,sintdx/1+x2  using 1
=sintdt=-cost +c=-cos(tan-1x)+c
```
7 years ago
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
```
7 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;
```
7 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).
```
7 years ago
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.

```
7 years ago
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