Question

How do you integrate `int 1/sqrt(-e^(2x) +49)dx` using trigonometric substitution?

Answer 1

`color(blue)(int 1/sqrt(49-e^(2x))dx=1/7*ln(7-sqrt(49-e^(2x)))-1/7x+C)`

Explanation:

The given `int 1/sqrt(49-e^(2x))dx`

Let `e^x=7*sin theta`
and `e^(2x)=49*sin^2 theta`
and `e^x dx=7*cos theta*d theta`
and `dx=(7*cos theta*d theta)/e^x=(7*cos theta*d theta)/(7*sin theta)=(cos theta*d theta)/(sin theta)`

Let us do the integration

`int 1/sqrt(49-e^(2x))dx=int 1/sqrt(49-49*sin^2 theta)*(cos theta*d theta)/(sin theta)`

`int 1/sqrt(49-e^(2x))dx=int 1/(sqrt(49)sqrt(1-sin^2 theta))*(cos theta*d theta)/(sin theta)`

`int 1/sqrt(49-e^(2x))dx=int 1/(7*sqrt(cos^2 theta))*(cos theta*d theta)/(sin theta)`

`int 1/sqrt(49-e^(2x))dx=int 1/(7*cos theta)*(cos theta*d theta)/(sin theta)`

`int 1/sqrt(49-e^(2x))dx=int 1/(7*cancelcos theta)*(cancelcos theta*d theta)/(sin theta)`

`int 1/sqrt(49-e^(2x))dx=1/7*int (d theta)/(sin theta)=1/7*int csc theta*d theta`

`int 1/sqrt(49-e^(2x))dx=1/7*ln(csc theta-cot theta)+C`

Now, we have to return it to the original variable x.

From our trigonometric substitution
`e^x=7*sin theta`

`sin theta=e^x/7` and our Right Triangle has opposite side`=e^x` and hypotenuse`=7` and adjacent side`=sqrt(49-e^(2x))`

Therefore,

`csc theta=7/e^x` and `cot theta=(sqrt(49-e^(2x)))/e^x`

And

`int 1/sqrt(49-e^(2x))dx=1/7*ln(csc theta-cot theta)+C`

`int 1/sqrt(49-e^(2x))dx=1/7*ln(7/e^x-(sqrt(49-e^(2x)))/e^x)+C`

`int 1/sqrt(49-e^(2x))dx=1/7*ln((7-sqrt(49-e^(2x)))/e^x)+C`

`int 1/sqrt(49-e^(2x))dx=1/7*[ln(7-sqrt(49-e^(2x)))-ln e^x]+C`

`int 1/sqrt(49-e^(2x))dx=1/7*ln(7-sqrt(49-e^(2x)))-1/7x+C`

God bless....I hope the explanation is useful.