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# integration by trig substitution [Solved!]

### My question

Chapter 8:Integration by Trig Sub.

Exercise 1: In the third step down the constant 16 is multiplied by the 1/2 in the integrand to arrive at 8. Why does 1/2 still exist in the next step after you integrate?

### Relevant page

8. Integration by Trigonometric Substitution

### What I've done so far

See above

X

Chapter 8:Integration by Trig Sub.

Exercise 1: In the third step down the constant 16 is multiplied by the 1/2 in the integrand to arrive at 8.  Why does 1/2 still exist in the next step after you integrate?
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

See above

## Re: integration by trig substitution

@Phinah: The first 1/2 is multiplied by the 16 outside to give 8 outside. The second 1/2 is the constant that comes from integrating cos 2x.

I've added another line in that solution which I hope makes it clearer what's happening. (You may need to refresh the page to see the change.)

X

@Phinah: The first 1/2 is multiplied by the 16 outside to give 8 outside. The second 1/2 is the constant that comes from integrating cos 2x.

I've added another line in that solution which I hope makes it clearer what's happening. (You may need to refresh the page to see the change.)