Ok math guru's my boyfriend put me up to this. The question is below thanks?

Question

the question is evaluate the integral from 0 to pi/8 of sec^2 2x / 3 + tan 2x dx

he says he got this 1/2 ln(1/3) but he's not sure if its done right.

thanks
- Jenny

Answer 1

note that d/dx(3+tan2x)=2sec^2 2x

Recall that int d/dxf(x)/f(x)=ln(f(x)) and use that directly
(or use the substitution u=3+tan2x)

Your boyfriend is close but made a small mistake.
Ignore the first poster's answer it is wrong!
(well it is mostly right, messed up the 3 in the substitution)

Answer 2

i imagine i know what you recommend... a million/ a million+a million/ a million+a million/ a million+a million/ ... obviously that's decrease than a million/a million = a million and is more beneficial than a million/(a million+a million) = a million/2 the 2d besides rt(a million+rt(a million+rt(a million+rt(a million+... back we see that's obviously more beneficial than rt1 = a million and obviously decrease than rt(a million+2) = rt(3) that's more beneficial or less a million.732 by the technique i visit call the sq. root basically "rt" for the first one... enable a million/a million+a million/a million+... = x now then shall we upload a million to both area a million + a million/a million+a million/a million+... = x+a million note some thing? now turn it over a million/ a million+ a million/a million+a million/... = a million/(x+a million) yet then you definately get what you all started with so x = a million/(x+a million) x^2 + x = a million x^2 + x - a million = 0 finished the sq. x^2 + x + a million/4 - 5/4 = 0 (x+a million/2)^2 - (rt5/2)^2 = 0 then if A^2 - B^2 = 0 A+B = 0 or A-B = 0 as a outcome x is -a million/2 plus or minus (rt5)/2 yet x is clearly an excellent huge form as a outcome x = ((rt5)-a million)/2 that's about 0.618 which seems logical pondering our first observations one problem finished! prob no.2 back, yet basically to lead away from confusion, i'm calling it y y = rt(a million+rt(a million+... back upload a million to both area y+a million = a million+rt(a million+rt(a million+... now take the rt of both area rt(y+a million) = rt(a million+rt(a million+... yet back this places us back to the starting up so... rt(y+a million) = y sq. both area y+a million = y^2 y^2 - y - a million = 0 finished the sq. y^2 - y +a million/4 - 5/4 = 0 (y-a million/2)^2 - ((rt5)/2)^2 = 0 interior a similar way we finish that y = a million/2 plus or minus (rt5)/2 y is clearly effective (and may be such, because that's a root!) as a outcome y = (a million + rt5)/2 which comes out about a million.618 consistent back! SOLVED! signed, Math guru :p

Answer 3

To do the integral

sub s = tan(2x) + 3
ds = 2sec(2x)^2 dx

so we have

=1/2 (int) 1/s ds

The int of 1/s = log(s)

so = log(s) / 2

Sub backk in the s = tan(2x) + 3

and you get

1/2 log(tan(2x)+3) then sub in the integral vals as given