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# 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

theta = arctan (0) = 0

theta = arctan (1) = .7853

X

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 thetas 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 thetas showed properly and also so the answer worked OK on a phone.