What is antiderivative of function f(x)=((sinx)raised to square+sinx)/(1+sinx+cosx)?calculate for limits x=0 and x=pi/2

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beckden | High School Teacher | (Level 1) Educator

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First notice that

` (1+cosx+sinx)^2=cos^2x+2cosxsinx+2cosx+sin^2x+2sinx+ 1`

 

 

`=cos^2x+sin^2x+2cosxsinx+2cosx+2sinx+1=1+2cosxsinx+2cosx+2sinx+1 `

 

`=2(1+cosxsinx+sinx+cosx)=2(1+cosx)(1+sinx) `

Now we multiply `(sinx(sinx+1))/(1+sinx+cosx)*(1+sinx+cosx)/(1+sinx+cosx)`

We get

`sinx(1+sinx)(1+sinx+cosx)/(1+sinx+cosx)^2` =sinx(1+sinx)

`(1+sinx+cosx)/(2(1+cosx)(1+sinx))`

`=sinx(1+sinx+cosx)/(2(1+cosx))`

`=sinx(1+cosx)/(2(1+cosx))+sin^2x/(2(1+cosx))`

`=1/2sinx+(1-cos^2x)/(2(1+cosx))`

`=1/2sinx+((1+cosx)(1-cosx))/(2(1+cosx))`

`=1/2sinx+(1-cosx)/2`

 

`=1/2sinx+1/2-1/2cosx `

The antiderivative of `1/2sinx+1/2-1/2cosx` is

`-1/2cosx+1/2x-1/2sinx`
So our answer is

`1/2x-1/2cosx-1/2sinx+C`

 

`int_0^(pi/2) (sin^2x+sinx)/(1+cosx+sinx)dx`

`=1/2(pi/2)-1/2cospi/2-1/2sinpi/2 - (1/2(0) - 1/2cos0-1/2sin0)`

`=pi/4 - 1/2(0) - 1/2(1) - (0 - 1/2(1) - 1/2(0))`

`= pi/4 - 1/2 + 1/2 = pi/4` Which is our answer

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