Question

Evaluate the integral (tanx+sinx)/secx dx from pi/6 to pi/3

Answer 1

\[\int\limits_{\pi/6}^{\pi/3} (tanx +sinx)/secx dx\]

Answer 2

why not rewrite as integral of sin(x) + (1/2)sin(2x) ?

Answer 3

simplify the expression 1st

\[\tan(x)/\sec(x) + \sin(x)/\sec(x) = sinx/\cos^2(x) + \tan(x)\]

Answer 4

i got to the point where sinx + sinxcosx dx

Answer 5

integral of sin(x)cos(x) is basically integral of sin(2x), although u-substituting u=sin(x) is also easy

Answer 6

split and do the integral of the sin(x) separately from the sin(x)cos(x)

Answer 7

so the problem becomes \[\int\limits_{0}^{\pi/3} \sin(x)/\cos^2(x) dx + \int\limits_{0}^{\pi/3} \tan(x) \]

Answer 8

i let u = cosx

Answer 9

so i get -cosx -cos^2x/2 +c

Answer 10

@campbell_st check your definition of 1/sec(x)

Answer 11

sec = 1/cos

Answer 12

right, so tan(x)/sec(x) = ?

Answer 13

sinx

Answer 14

and tan = sin/cos to tan/sec = sin/cos x 1/cos or sin/cos^2

Answer 15

@Lammy -- you typed into OpenStudy "Evaluate the integral (tanx+sinx)/secx dx from pi/6 to pi/3" I cutpasted it into www.wolframalpha.com "Evaluate the integral (tanx+sinx)/secx dx from pi/6 to pi/3" and got 1/4 (-1+2 sqrt(3))

Answer 16

@campbell_st -- no that should be tan/sec = (sin/cos) / (1/cos) or sin

Answer 17

lol... sorry about the boundary mis-typing...

Answer 18

for the 1st integral let u = cos x then du = - sin(x) dx

so the integral is \[\int\limits_{?}^{} du/u^2\]

Answer 19

you'll need to find the boundary values...

Answer 20

the boundary is 1/2 and sqrt(
3)/2

Answer 21

oops it should be \[- \int\limits_{\sqrt{3}/2}^{1/2} du/u^2 \]

the other part is \[\int\limits_{\pi/6}^{\pi/3} \tan(x) dx\]

so

Answer 22

ok i think i figure it out

Answer 23

i got sqrt3/2 - 1/4 same as wolfram