Question

How do you integrate `int (x^2-1)/sqrt(x^2+9)dx` using trigonometric substitution?

Answer 1

`int(x^2-1)/sqrt(x^2+9)dx=x/2sqrt(x^2+9)-11/2ln|x+sqrt(x^2+9)|+C`

Explanation:

Here ,

`int(x^2-1)/sqrt(x^2+9)dx=int(x^2+9-10)/sqrt(x^2+9)dx`

`int(x^2-1)/sqrt(x^2+9)dx=intsqrt(x^2+9)dx-10int1/sqrt(x^2+9)dx`

`int(x^2-1)/sqrt(x^2+9)dx=I-10ln|x+sqrt(x^2+9)|color(red)(...to(A)`

Now, `I=intsqrt(x^2+9)dx`

`:.I=intsqrt(x^2+9)*1dx`

Using Integration by parts:

`I=sqrt(x^2+9)int1dx-int(1/(2sqrt(x^2+9))(2x)int1dx)dx`

`I=sqrt(x^2+9)*x-intx/sqrt(x^2+9)xdx`

`=xsqrt(x^2+9)-intx^2/sqrt(x^2+9)dx`

`=xsqrt(x^2+9)-int(x^2+9-9)/sqrt(x^2+9)dx`

`I=xsqrt(x^2+9)-intsqrt(x^2+9)dx+int9/sqrt(x^2+9)dx`

`I=xsqrt(x^2+9)-I+9ln|x+sqrt(x^2+9)|+c`

`2I=xsqrt(x^2+9)+9ln|x+sqrt(x^2+9)|+c`

`I=x/2sqrt(x^2+9)+9/2ln|x+sqrt(x^2+9)|+c/2`

From eqn `color(red)((A)` we have

`int(x^2-1)/sqrt(x^2+9)dx=x/2sqrt(x^2+9)+9/2ln|x+sqrt(x^2+9)|`

`color(white)(............................................)-10ln|x+sqrt(x^2+9)|+c/2`

`int(x^2-1)/sqrt(x^2+9)dx=x/2sqrt(x^2+9)-11/2ln|x+sqrt(x^2+9)|+C`