Question

integral x tan^2 x dx
solve using parts .. please help

Answer 1

Let u = x dv = sec^2x dx
du = dx v = tan x

so by "integration by parts", you get

x tan(x) - Integral of (tan x) dx

= x (tan x) + ln/cos x/ + C

Keep in mind that the integral of tan x is -ln/cos x/

Answer 2

i thought the integral of tan x was ln/secx/+c

Answer 3

ok, that will also work..either one is correct.

Answer 4

Either one is correct; as they will both be correct after checking with differentiation.

Answer 5

thnx :)

Answer 6

so you will have x tan(x) - ln/sec x/ + C
@karid

Answer 7

but wouldn't dv = tan^2 x

Answer 8

@karid

BTW, if you ever forgot the integral of tan x...just write tan x as sin x/cos x...let u = cos x, du = -sinx dx

so you get the integral of tan x dx = -ln/cos x/ + C

Answer 9

@karid

You are absolutely right...I thought it said sec^2(x)

so tan^2(x) = sec^2(x) - 1

so integrate that...tan x - x

sorry for the oversight.

Answer 10

its ok ....thnx that makes more sense ..

Answer 11

From Pythagorean Identity,

1 + tan^2(x) = sec^(2)x, so tan^2(x) = sec^2(x) - 1

@karid

Answer 12

i know how to do the rest i forgot that i could change tan^2 x to sec^2 - 1 .. :)

Answer 13

OK. All is cool.