phinah 10 Apr 2018, 10:44
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?
8. Integration by Trigonometric Substitution
See above
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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
Murray 11 Apr 2018, 02:48
@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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@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.)