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IntMath forum | Methods of Integration

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.)

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