Question

plseeeeeeeeee i need it urgentlyyyyyyyyyyyyyy
need it very badly...very urgent

Answer 1

Can't do integrals; sorry.

Answer 2

who can do it...plse tell me

Answer 3

UnkleRhaukus is the best in the business I think.

Answer 4

i really dont think i can help here

Answer 5

An indefinite integral?
Why don't you just differentiate the right-hand side and hope for the best? :D

Answer 6

doodling...
\[\large 1 + \tan(x)\tan(x+\theta) = 1+\tan(x)\left[\frac{\tan(x)+\tan(\theta)}{1-\tan(x)\tan(\theta)}\right]\]

Answer 7

Just to spare myself from confusion, sort of, I'm letting
\[\large p = \tan(\theta)\]
\[\large 1 + \frac{\tan^2(x) + p\tan(x)}{1-p\tan(x)}\]

Answer 8

\[\large =\frac{1-p\tan(x)+\tan^2(x)+p\tan(x)}{1-p\tan(x)}\]

Answer 9

\[\large \frac{1+\tan^2(x)}{1-p\tan(x)}\]\[\large \frac{\sec^2(x)}{1-p\tan(x)}\]

Answer 10

This looks much more palatable, don't you think?
@msingh
@ParthKohli
@UnkleRhaukus
:D

Answer 11

i think i got it

Answer 12

I was just getting to the fun part >.>
Oh well, good on you :D

Answer 13

From where @terenzreignz left it, the integration should be easy; let u=1-ptanx and doo-wop that thing.

Answer 14

doo-wop :D

Answer 15

That thing, that thing, that thiiiiiiing. Like Lauryn Hill.