`int 6/sqrt(10x-x^2) dx` Find the indefinite integral

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`int6/sqrt(10x-x^2)dx`

Take the constant out ,

`=6int1/sqrt(10x-x^2)dx`

Let's rewrite the denominator by completing the square,

`=6int1/(sqrt(-(x-5)^2+25))dx`

Now apply integral substitution: `u=x-5`

`du=1dx`

`=6int1/sqrt(-u^2+25)dx`

`=6int1/sqrt(5^2-u^2)du`

Again apply integral substitution: `u=5sin(v)`

`=>du=5cos(v)dv`

`=6int(1/sqrt(5^2-5^2sin^2(v)))5cos(v)dv`

`=6int(5cos(v))/(sqrt(5^2)sqrt(1-sin^2(v)))dv`

Use the trigonometric identity: `cos^2(x)=1-sin^2(x)`

`=6int(5cos(v))/(5sqrt(cos^2(v)))dv`

`=6int(5cos(v))/(5cos(v))dv` , assuming `cos(v)>=0`

`=6int1dv`

`=6v` 

substitute back `v=arcsin(u/5)` and `u=(x-5)`  

`=6arcsin((x-5)/5)`

Add a constant C...

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`int6/sqrt(10x-x^2)dx`

Take the constant out ,

`=6int1/sqrt(10x-x^2)dx`

Let's rewrite the denominator by completing the square,

`=6int1/(sqrt(-(x-5)^2+25))dx`

Now apply integral substitution: `u=x-5`

`du=1dx`

`=6int1/sqrt(-u^2+25)dx`

`=6int1/sqrt(5^2-u^2)du`

Again apply integral substitution: `u=5sin(v)`

`=>du=5cos(v)dv`

`=6int(1/sqrt(5^2-5^2sin^2(v)))5cos(v)dv`

`=6int(5cos(v))/(sqrt(5^2)sqrt(1-sin^2(v)))dv`

Use the trigonometric identity: `cos^2(x)=1-sin^2(x)`

`=6int(5cos(v))/(5sqrt(cos^2(v)))dv`

`=6int(5cos(v))/(5cos(v))dv` , assuming `cos(v)>=0`

`=6int1dv`

`=6v` 

substitute back `v=arcsin(u/5)` and `u=(x-5)`  

`=6arcsin((x-5)/5)`

Add a constant C to the solution,

`=6arcsin((x-5)/5)+C`

 

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