Question

How do I find the integral of `int sqrt((5-x^2))`?

Answer 1

`I=x/2sqrt(5-x^2)+5/2arcsin(x/sqrt5)+c`

Explanation:

Here ,

`I=intsqrt(5-x^2)*1dx`

Using Integration by parts:

`I=sqrt(5-x^2)int1dx-int(d/(dx)(sqrt(5-x^2))int1dx)dx`

`I=sqrt(5-x^2)xx x-int1/(2sqrt(5-x^2)) (-2x)(x)dx`

`I=xsqrt(5-x^2)-int(-x^2)/sqrt(5-x^2)dx`

`I=xsqrt(5-x^2)-int((5-x^2)-5)/sqrt(5-x^2)dx`

`I=xsqrt(5-x^2)-int(5-x^2)/sqrt(5-x^2)dx+int5/sqrt(5-x^2)dx`

`I=xsqrt(5-x^2)-intsqrt(5-x^2)dx+5int1/sqrt((sqrt5)^2-x^2)dx`

`I=xsqrt(5-x^2)-I+5arcsin(x/sqrt5)+c`

`I+I=xsqrt(5-x^2)+5arcsin(x/sqrt5)+c`

`2I=xsqrt(5-x^2)+5arcsin(x/sqrt5)+c`

`I=x/2sqrt(5-x^2)+5/2arcsin(x/sqrt5)+c`