Find integral sqrt (x^2 + 1) using trigonometric substitution [Solved!]

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Find  `int sqrt (x^2 + 1) dx` with limits of integration from 0 to 1 using Trigonometric Substitution.

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

I used `x = tan theta` and `dx = sec^2 theta d theta.`

Replacing `x` and `dx` gives `sqrt (tan^2 theta + 1)  sec^2 theta` `= sqrt (sec^2 theta)  sec ^2 theta` `= sec ^3 theta d theta`.

I looked up the integral of `sec ^3 theta` and converted `tan,` `sec,` `sin,` and cos in the formula by drawing a triangle based on `x = tan theta` and arrived at `.57155` and not `1.15.`

X

@phinah: Good on you for using the math entry system! (I tidied up some of the math expressions.)

Just a small (but important) point - don't miss out the "`d theta`" parts in your second line. It should be:

Replacing `x` and `dx` gives `sqrt (tan^2 theta + 1)  (sec^2 theta) d theta` `= sqrt (sec^2 theta)  (sec ^2 theta) d theta` `= sec ^3 theta d theta`.

Now, to give you a hint about why your final number is not correct (I'm guessing `1.15` comes from the Answers in your text book, right?).

When we change `x` to `tan theta` and `x` goes from `0` to `1`, what will `theta`'s lower and upper values be?

X

In radians:

`theta = arctan (0) = 0`

`theta = arctan (1) = .7853`

X

Also this is not a textbook exercise.  This is from your website, the question at the end of the Integration chapter, section 4: The Definite Integral. You state it can be solved via Trigonometric Substitution but you solved it using the <a href="https://www.intmath.com/integration/5-trapezoidal-rule.php">Trapezoidal Rule in section 5</a>. Thanks.

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Ahh, I see. Thanks for giving me the context.

Your lower and upper values are correct. Now substitute those in the integral of `sec^3 theta` you found from the table.

X

Leaving it in terms of theta:

`int sec^3 theta` `d theta` from `0` to `.7853`

According to Wolfram, the integral formula is

`1/2[tan theta sec theta` `{: - ln (cos {:theta/2:} - sin {:theta /2:})` `{: + ln (sin {:theta/2:} + cos {:theta/2:})]`

Therefore, `1/2 [tan .7853\ sec .7853 ` `{:- ln (cos .7853/2 - sin .7853/2)` `{: + ln (sin .7853/2 + cos .7853/2)] ` `- 1/2 [0 - ln (1-0) + ln (0+1) ]` 

`= 1/2[tan .7853\ sec .7853` `{: - ln (cos .7853/2 - sin .7853/2)` `{: + ln (sin .7853/2 + cos .7853/2)]` ` - [0]`

`=1.15 `

which is the web site's answer using the Trapezoidal Rule!

Thank you for the guidance.

Note: concerning how `theta` is written above, I wrote it out in Word.

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OK - looks good.

BTW, the "theta" symbols didn't show because Word uses a different set of fonts. It's always best to type the math directly in the text box, for best results. I edited your answer so the `theta`s showed properly and also so the answer worked OK on a phone.

X

Got it.  Thanks.

After finding the integral in terms of `theta` you state that we can change back to x or leave it in terms of `theta`. Above, I left it in terms of theta.

Part Two is to change it back to x. Trying to figure out why I did not arrive at the same answer like I should have.

What I've done so far

Changing back to x:

`int sec^3 x`  `d x` =

1/2 [tan x sec x  ln (cos (x/2)  sin (x/2)) + ln (sin (x/2) + cos (x/2))]

Referencing x = tan `theta`: my right triangle is x opposite `theta`, 1 adjacent to `theta`, and hypotenuse = `sqrt (x^2 + 1)`.

Substituting sin, cos, tan, and sec from the triangle into the integral from 0 to 1:

1/2 [(x)(`sqrt (x^2+1)`)  ln (1/`sqrt (x^2+1)`  
x/`sqrt (x^2+1)`) + ln (x/`sqrt (x^2+1)` + 1/`sqrt (x^2+1)`)


= 1/2 [(1)(`sqrt 2`)  ln (.8944 - .4472) + ln (.4472 + .8944)]  1/2 [(0)(`sqrt 1`)  ln (1  0) + ln (0 + 1)] 

= 1/2 [`sqrt 2`  ln(.4472) + ln(1.3416)]  [ 0 ]

= 1.26

Thanks in advance for the additional guidance.

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But we can't just do this (trade `theta`s for `x`s)!

`int sec^3 x d x`

`=1/2[tan x sec x - ln (cos (x/2) - sin (x/2))` `{: + ln (sin (x/2) + cos (x/2))]`


The expressions in `theta` need to stay in theta, then we substitute back into `x.` You kind of did bits of that eventually, but it's important we write it correctly.

This is what we had before:

`int sec^3 theta d theta`

`=1/2[tan theta sec theta` `{: - ln (cos {:theta/2:} - sin {:theta /2:})` `{: + ln (sin {:theta/2:} + cos {:theta/2:})]`

We integrated it in terms of `theta` so we need to have the result in terms of `theta.`

We will use `tan theta = x` and `sec theta = sqrt(x^2+1)` from the triangle you constructed.

But for `cos(theta/2)` we need to use the <a href="https://www.intmath.com/analytic-trigonometry/4-half-angle-formulas.php">Half Angle Formula</a>:

`cos(theta/2) = sqrt((1 + cos theta)/2)`

Expressing the RHS in terms of `x` we need to use (from the triangle) `cos theta = 1/(sqrt(x^2+1))` and this will give us:

`cos(theta/2) = sqrt((1 + 1/(sqrt(x^2+1)))/2)`

Do you think you can proceed from there?

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The half-angle formula for sine is similar to that for cosine only we subtract in the numerator.

Substituting the limits of integration:

`1/2 [1( sqrt 2) - ln(.924 - .383)` `{: + ln(.383 + .924)] ` `- 1/2 [0(1) ` `- ln (1-0)` `{: + ln (0 + 1)]` ` = 1.148 ~~ 1.15`

The answer given using the Trapezoidal Rule!

Thanks!