Question

Did I do this right?

Answer 1

\[\int\limits \tan ^{2}x dx\]

>rewrote as \[\int\limits \frac{ sinx }{ 1+cosx } dx\]

Answer 2

@amistre64

Answer 3

>set u=cosx
>du/dx=sinx
>du=sinxdx
>sub in
>\[\int\limits \frac{ 1 }{ 1+u } du\]

Answer 4

>take antiderivative
> \[\ln (1+u)\]

Answer 5

>sub cosx back for u
>final answer is ln(1+cosx)

Is this right?

Answer 6

actually ln (1+cosx) +C ofc

Answer 7

@zepdrix @.Sam.

Answer 8

@ganeshie8

Answer 9

@Zarkon

Answer 10

Its kinda hard to see....

Answer 11

yea sorry about that

Answer 12

@tester97

Answer 13

anyone?

Answer 14

so is my answer correct?

Answer 15

I can answer if you make it more clear to see.