Question

How do you integrate `int (25-x^2)/sqrt(x^2+4)dx` using trigonometric substitution?

Answer 1

`27lnabs(x+sqrt(x^2+4))-1/2xsqrt(x^2+4)+C`

Explanation:

We will want to use the substitution `x=2tantheta`. This implies that `dx=2sec^2thetacolor(white).d theta`.

`I=int(25-x^2)/sqrt(x^2+4)dx=int(25-4tan^2theta)/sqrt(4tan^2theta+4)(2sec^2thetacolor(white).d theta)`

Note that `sqrt(4tan^2theta+4)=2sqrt(tan^2theta+1)=2sectheta`. The fact that we use `tan^2theta+1=sec^2theta` to simplify the square root was the reason to choose the substitution `x=2tantheta` in the first place.

`I=int(25-4tan^2theta)/(2sectheta)(2sec^2theta)d theta=int(25-4tan^2theta)(sec theta)d theta`

Expanding and writing `tan^2thetasectheta` in terms of `sintheta` and `costheta`:

`I=25intsecthetacolor(white).d theta-4intsin^2theta/cos^3thetad theta`

Rewrite `sin^2theta` as `1-cos^2theta`:

`I=25intsecthetacolor(white).d theta-4int(1-cos^2theta)/cos^3thetad theta`

`I=25intsecthetacolor(white).d theta-4int(sec^3theta-sectheta)d theta`

`I=29intsecthetacolor(white).d theta-4intsec^3thetacolor(white).d theta" "" "color(red)((star))`

We will tackle `intsec^3thetacolor(white).d theta` with integration by parts. Since `sec^3theta=sectheta(sec^2theta)`, and `sec^2theta` can easily be integrated, let:

`{(u=sectheta,=>,du=secthetatanthetacolor(white).d theta),(dv=sec^2thetacolor(white).d theta,=>,v=tantheta):}`

Then, using `intudv=uv-intvdu`:

`I=29intsecthetacolor(white).d theta-4(secthetatantheta-intsecthetatan^2thetacolor(white).d theta)`

`I=29intsecthetacolor(white).d theta-4secthetatantheta+4intsecthetatan^2thetacolor(white).d theta`

Now letting `tan^2theta=sec^2theta-1`:

`I=29intsecthetacolor(white).d theta-4secthetatantheta+4intsectheta(sec^2theta-1)d theta`

`I=25intsecthetacolor(white).d theta-4secthetatantheta+[4intsec^3thetacolor(white).d theta]`

Returning to `color(red)((star))`, we see that `4intsec^3thetacolor(white).d theta=29intsecthetacolor(white).d theta-I`:

`I=25intsecthetacolor(white).d theta-4secthetatantheta+[29intsecthetacolor(white).d theta-I]`

Combining and adding `I` to both sides:

`2I=54intsecthetacolor(white).d theta-4secthetatantheta`

`I=27intsecthetacolor(white).d theta-2secthetatantheta`

The antiderivative of `sectheta` is commonly known:

`I=27lnabs(tantheta+sectheta)-2secthetatantheta`

Our original substitution was `x=2tantheta`, which implies that `tantheta=x/2`.

From this we can say that `sectheta=sqrt(tan^2theta+1)=sqrt(x^2/4+1)=1/2sqrt(x^2+4)`. Then:

`I=27lnabs(x/2+1/2sqrt(x^2+4))-2(1/2sqrt(x^2+4))x/2+C`

Factoring `1/2` from the natural logarithm, it can be removed from the logarithm using `ln(a/b)=ln(a)-ln(b)`, and the resultant `-ln(2)` will be absorbed into the constant of integration.

`I=27lnabs(x+sqrt(x^2+4))-1/2xsqrt(x^2+4)+C`

Answer 2

`int frac{25-x^2}{sqrt(x^2+4)}dx`

`=27ln|[x+sqrt[x^2+4]]/2|-(xsqrt[x^2+4])/2+C`

Explanation:

`int frac{25-x^2}{sqrt(x^2+4)}dx`

`=int frac{25-x^2}{sqrt[(4)(x^2/4+1)]}`

`=1/2int frac{25-x^2}{sqrt((x/2)^2+1)}`

Recognize trig substitution in denominator from radical, so list various useful trigonometric ratios with this triangle.

`color(green)(tantheta=x/2)`
`color(green)(x=2tan theta)`
`color(green)(dx=2sec^2theta (d theta))`
`color(green)(25-x^2=25-4tan^2theta)`

`color(green)(sectheta=sqrt[(x/2)^2+1])`

`=1/2intfrac{25-4tan^2theta}{sectheta}(2sec^2theta)(d theta)`

`=1/2intfrac{25-4tan^2theta}{cancel(sectheta)}(2cancel(sectheta)sectheta)(d theta)`

`=int[25sectheta-4secthetatan^2theta] d theta`

`=25int(sectheta )d theta-4int(secthetatan^2theta )d theta`

`=25ln|sectheta+tantheta|-4int(secthetatan^2theta)d theta`

Now, we're getting close to the end - we have to simplify `color(red)(int(secthetatan^2theta)d theta)`
Use integration by parts:
`color(red)(u=tantheta)`
`color(red)(du=sec^2theta)`
`color(red)(v=sectheta)`
`color(red)(dv=secthetatantheta)`

`color(red)(int(secthetatan^2theta)d theta=secthetatantheta-int(sec^3theta) d theta)`
`color(red)(=secthetatantheta-int(sectheta)(1+tan^2theta)d theta)`
`color(red)(=secthetatantheta-int(sectheta)d theta-int(secthetatan^2theta)d theta)`
`color(red)(int(secthetatan^2theta)d theta=secthetatantheta-ln|sectheta+tantheta|-int(secthetatan^2theta)d theta)`
`color(red)(2intsecthetatan^2theta d theta=secthetatantheta-ln|sectheta+tantheta|)`
`color(red)(intsecthetatan^2theta d theta=1/2secthetatantheta-1/2ln|sectheta+tantheta|+C)`

`=25ln|sectheta+tantheta|-4[1/2secthetatantheta-1/2ln|sectheta+tantheta|]+C`

`=25ln|sectheta+tantheta|-2secthetatantheta+2ln|sectheta+tantheta|+C`

`=27ln|sectheta+tantheta|-2secthetatantheta+C`

Lastly, re-substitute the trig ratios:
`=27ln|sqrt[(x/2)^2+1]+x/2|-2(x/2)sqrt[(x/2)^2+1]+C`

`=27ln|x/2 + sqrt[(x/2)^2+1]|-xsqrt[(x/2)^2+1]+C`

Change the square root notation back to the original form:
`=27ln|[x+sqrt[x^2+4]]/2|-(xsqrt[x^2+4])/2+C`