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INTEGRATION [Solved!]

My question

Section 8, Integration by Trig Sub, Exercise 3:

We are not clear as to why you have + 1 in the solution.

Relevant page

8. Integration by Trigonometric Substitution

What I've done so far

Because the integral is of the form `x^2 - a^2` after completing the square, we used x = a `sec theta` with a = 1. Setting up the triangle, x is the hypotenuse, 1 is the adjacent side, and `sqrt (x^2+2x)` is the opposite side.

Therefore secant = x and tan = `sqrt (x^2+2x)` should be substituted into the integral of sec x to obtain the final answer.

X

Section 8, Integration by Trig Sub, Exercise 3:


We are not clear as to why you have + 1 in the solution.
Relevant page

<a href="https://www.intmath.com/methods-integration/8-integration-trigonometric-substitution.php">8. Integration by Trigonometric Substitution</a>

What I've done so far

Because the integral is of the form `x^2 - a^2` after completing the square, we used x = a `sec theta` with a = 1.  Setting up the triangle, x is the hypotenuse, 1 is the adjacent side, and `sqrt (x^2+2x)` is the opposite side.

Therefore secant = x and tan =  `sqrt (x^2+2x)` should be substituted into the integral of sec x to obtain the final answer.

Re: INTEGRATION

I don't know why this question was missed. So sorry!

I gather you mean the "+1" in the line

"The triangle in this case starts with `x+1= sec theta + 1`"

Right?

Because I substituted `u=x+1` and we chose `u = sec theta` earlier in the problem, then we'll have:

`u = x+1 = sec theta + 1"

Looking at your solution, I agree with using `a sec theta`, but it should be like this:

Because the integral is of the form `X^2 - a^2` after completing the square (with `X = x+1`), we use

`X = a sec theta` with `a = 1.` That is, `x+1 = sec theta.`

The triangle will now be as I have in the solution.

Hope it helps.

X

I don't know why this question was missed. So sorry!

I gather you mean the "+1" in the line 

"The triangle in this case starts with `x+1= sec theta + 1`"

Right?

Because I substituted `u=x+1` and we chose `u = sec theta` earlier in the problem, then we'll have:

`u = x+1 = sec theta + 1"

Looking at your solution, I agree with using `a sec theta`, but it should be like this:

Because the integral is of the form `X^2 - a^2` after completing the square (with `X = x+1`), we use

`X = a sec theta` with `a = 1.` That is, `x+1 = sec theta.`  

The triangle will now be as I have in the solution.

Hope it helps.

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