# integrate sinx dx / (1+sinx)^1/2

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### 1 Answer

`intsinx/sqrt(1+sinx)dx `

Let, `u=sinx, cosx=sqrt(1-u^2)` (draw a right triangle to get support for this value of cosx)

`du=cosxdx`

`dx=(du)/sqrt(1-u^2)`

The integral then takes the form,

`intu/(sqrt(1+u)sqrt(1-u^2))du `

=`intu/(sqrt((1+u)(1-u)(1+u)))du `

`=intu/((1+u)sqrt(1-u))du`

Make a second substitution as `sqrt(1-u)=t `

`u=1-t^2`

`du=-2tdt`

The integral now becomes

`int((1-t^2)(-2t))/((2-t^2)t)dt `

`=-2int(1-t^2)/(2-t^2)dt`

`=-2int((2-t^2)/(2-t^2)-1/(2-t^2))dt`

`=-2int(1-1/(2-t^2))dt`

Integration by decomposing into partial fraction yields

`=-1/2(4t+sqrt2log(sqrt2-t)-sqrt2log(sqrt2+t))+C`

Putting back the value of t,

`=-1/2(4sqrt(1-u)+sqrt2log(sqrt2-sqrt(1-u))-sqrt2log(sqrt2+sqrt(1+u)))+C`

Finally, putting back the value of u gives,

`=-1/2(4sqrt(1-sinx)+sqrt2log(sqrt2-sqrt(1-sinx))-sqrt2log(sqrt2+sqrt(1+sinx)))+C` Where, C is the constant of integration.

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