Question

How do you integrate `int (e^x-1)/sqrt(e^(2x) -16)dx` using trigonometric substitution?

Answer 1

`-i(arcsin(e^x/4)+1/4ln(|(4+sqrt(16-e^(2x)))/e^x|))+c`

Explanation:

`e^x=4sinΘ`
`x=ln(4sinΘ)`
`dx=1/(4sinΘ)*4cosΘdΘ`
`dx=cosΘ/sinΘdΘ`

`int(e^x-1)/(sqrt(e^(2x)-16)` `dx`

`int(4sinΘ-1)/(sqrt((4sinΘ)^2-16)` `*cosΘ/sinΘdΘ`

`int(4sinΘ-1)/(sqrt(16sin^2Θ-16))` `*cosΘ/sinΘdΘ`

`int(4sinΘ-1)/(sqrt(16)*sqrt(sin^2Θ-1)` `*cosΘ/sinΘdΘ`

`int(4sinΘ-1)/(4*sqrt(-cos^2Θ)` `*cosΘ/sinΘdΘ`

`1/iint(4sinΘ-1)/(4cosΘ` `*cosΘ/sinΘdΘ`

`-iint(4sinΘ-1)/(4sinΘ)dΘ`

`-iint((4sinΘ)/(4sinΘ)-(1)/(4sinΘ))dΘ`

`-iint1dΘ+iint1/4cscΘdΘ`

`-iΘ + i/4ln|cscΘ+cotΘ| + c`

`color(red)( bar( ul( | color(white)(a/a) color(black)( -iarcsin(e^x/4)-i/4ln(|(4+sqrt(16-e^(2x)))/e^x|)+c, c in RR ) color(white)(a/a) | )))`

Note 1: We set `e^x` as a form of `sinΘ` because the `e^x` term in the root in the denominator was positive and the constant term was negative.

Note 2: `intcscΘ`=`-ln|cscx+cotx|+c`

Note 3: You get the expression for `Θ` from rewriting the original equation, and you can solve for `cscΘ` and `cotΘ` using a right triangle.

Note 4: There are probably ways to simplify my final answer, but I just kept it like that.

Answer 2

`I=int(e^x-1)/sqrt(e^(2x)-16)dx`
Let `X=e^x`

`x=ln(X)`

`dx=(dX)/X`

So:

`I=int(X-1)/(Xsqrt(X^2-16))dX`

`=int(cancel(X))/(cancel(X)sqrt(X^2-16))dX-int1/(Xsqrt(X^2-16))dX`

Now let `X=4sec(theta)`

`dX=4sec(theta)tan(theta)d theta`

So :

`I=int(4sec(theta)tan(theta))/sqrt((4sec(theta))^2-16)d theta-int(cancel(4sec(theta))tan(theta))/(cancel(4sec(theta))sqrt((4sec(theta))^2-16))d theta`

`=int(4sec(theta)tan(theta))/sqrt(16(sec(theta)^2-1))d theta-int(tan(theta))/sqrt(16(sec(theta)^2-1))d theta`

Because `sec(theta)^2-1=tan(theta)^2`,

`I=int(cancel(4)sec(theta)cancel(tan(theta)))/(cancel(4tan(theta)))d theta-intcancel(tan(theta))/(4cancel(tan(theta)))d theta`

`=intsec(theta)d theta-1/4int1d theta`

Now, we have `intsec(theta)d theta=ln(|sec(theta)+tan(theta)|)`

Here is a proof if you need the explanation.

So :
`I=ln(|sec(theta)+tan(theta)|)-1/4theta`

Because `theta=sec^(-1)(X/4)=sec^(-1)(e^x/4)`, and `tan(sec^(-1)(u))=sqrt(u^2-1)`, where `u` is a function (here is a proof if you need it.),

`I=ln(|e^x/4+sqrt(e^(2x)/16-1)|)-1/4sec^(-1)(e^x/4)+C`, `C in RR`.

\0/ Here's our answer !