# 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

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)]  - 

=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)]  - 

=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.

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

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

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.

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

But we can't just do this (trade thetas for xs)!

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 Half Angle Formula:

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?

X

But we can't just do this (trade thetas for xs)!

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?

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

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!

X

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!

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