Question

If we let x=tan theta, then definite integral on interval [1,rt3] of sqrt(1+x^2) is equivalent to...

Answer 1

If we let\[x=\tan \theta\]
then
\[\int\limits_{1}^{\sqrt{2}}\sqrt{1+x ^{2}}dx\]
is equivalent to...

Answer 2

sin^2(x) + cos^2(x) = 1

turns into

tan^2(x) + 1 = sec^2(x)

after you divide everything by cos^2(x)

Answer 3

tan^2(x) + 1 = sec^2(x)

sec^2(x) = tan^2(x) + 1

sec(x) = sqrt( tan^2(x) + 1 ) , where sec(x) > 0

Answer 4

so then it's just integral of secx?

Answer 5

yes

Answer 6

you'll have to change the limits of integration

Answer 7

but how do i adjust the limits if tanx=rt3 has 2 solutions?

Answer 8

i know that the bottom limit is adjusted to pi/4

Answer 9

since sec(x) > 0, this means cos(x) > 0

hmm what to do with tan(x)

Answer 10

oh wait in my answer choices only pi/3 is shown

Answer 11

well if tan(theta) = sqrt(3), and we just look at the principal value, then

theta = arctan(sqrt(3)) = pi/3

Answer 12

so it has to be pi/3

Answer 13

good, so you're going from pi/4 to pi/3

Answer 14

yup

Answer 15

thank you!

Answer 16

np