MATH INTEGRAL Help?

Question

integrate 11sinx+12tanx


i have no idea what to do with the tan.. i know i should do something with a u ?? but .. i duno.

Answer 1

integral of sin x is -cos x

(integral) tan x dx = (integral) sin x /cos x dx
set
u = cos x.
then we find
du = - sin x dx

substitute du=-sin x, u=cos x

(integral) sin x /cos x dx = - (integral)
(-1) sin x dx/ cos x

= - (integral) du/ u
Solve the integral

= - ln |u| + C

substitute back u=cos x

= - ln |cos x| + C


Answer -11cos x -12ln |cos x| + C

Answer 2

Let's just do ∫ tan x dx,
since that seems to be the part baffling you.
Write tan x = sin x /cos x
and notice that sin x is minus the derivative of cos x.
Thus ∫ sin x/cos x dx = - ln(cos x) + C = ln(sec x) + C,
since the integral is in the form ∫ -du/u,
with u = cos x.
BTW: Jenn gave you the derivative of tan x,
not its integral.

Answer 3

well there are a couple of ways to do it...

Below is one way...

S(11sinx+12tanx)dx = S(11sinx+12(sinx/cosx))dx...

this is because tanx = sinx/cosx

now you can distribute out sinx...

it becomes,

S(11+12/cosx)sinxdx ... now you can use "u" or substitution

u = cosx and du = -sinxdx

so, the integral becomes...

-S(11+12/u)du... integrate and you get

-11u-12ln(u) + C (C being some constant)

so you sub back in cosx... and the answer is

S(11sinx+12tanx)dx = -(11cosx + 12ln(cosx)) + C = -11cosx-12ln(cosx) + C

also, -12ln(cosx) = 12ln(1/cosx) = 12ln(secx)... so the answer can be...

12ln(secx)-11cosx + C

PS The guy below me means "jenn" not me :)

Answer 4

write tan(x) as sin(x)/cos(x)
so we have:
sin(x)dx/cos(x)
u=cos(x)
du=-sin(x)dx
substitute in to get
-du/u
the integral of du/u is ln[abs(u)]
so -du/u -> -ln[abs(u)]
now substitute the x's back in to get:
-ln[abs(cos(x))]
that covers the tan(x) part

Answer 5

int. tanx = (- ln cosx )

*Edit : biowolf98 is right

Answer 6

-11cosx + -12secx^2