IntMath Home » Forum home » Methods of Integration » Find integral sqrt (x^2 + 1) using trigonometric substitution

IntMath forum | Methods of Integration

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

My question

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

Relevant page

8. Integration by Trigonometric Substitution

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

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.`

Re: Find integral sqrt (x^2 + 1) using trigonometric substitution

@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

@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?

Re: Find integral sqrt (x^2 + 1) using trigonometric substitution

In radians:

`theta = arctan (0) = 0`

`theta = arctan (1) = .7853`

X

In radians:

`theta = arctan (0) = 0`

`theta = arctan (1) = .7853`

Re: Find integral sqrt (x^2 + 1) using trigonometric substitution

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 Trapezoidal Rule in section 5. Thanks.

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.

Re: Find integral sqrt (x^2 + 1) using trigonometric substitution

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

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.

Re: Find integral sqrt (x^2 + 1) using trigonometric substitution

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.

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.

Re: Find integral sqrt (x^2 + 1) using trigonometric substitution

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

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.

Reply

You need to be logged in to reply.