Question

how do you integrate 1-(tan(x))^2 ????? HElP

Answer 1

\[\int\limits 1-(\tan(x))^2\]

Answer 2

\[\int 1-tan^2x\]

Answer 3

yes

Answer 4

sorry with respect to x

Answer 5

You can separate it into two integrals:

\[\large \int 1dx -\int tan^2(x)dx\]

Answer 6

Hahahahhaha I have been at it toooooo long I should have seen that it is easy!!!!

Answer 7

thanks

Answer 8

Welcome :)

Answer 9

ok @zepp I am now scratching my head because I know how to integrate tanx but not \[\tan ^{2}x\]

Answer 10

Okay, first recall the trig identity: \(\large sin^2(x)+cos^2(x)=1\).

Answer 11

Divide everything by \(cos^2x\):
\[\large \frac{sin^2x}{cos^2x}+\frac{cos^2x}{cos^2x}=\frac{1}{cos^2x}\\\large tan^2x+1=sec^2x\\\large tan^x=sec^2x\]

Answer 12

Now we have \[\large \int1-tan^2xdx=\int1dx-\int tan^2xdx\\\large=\int1dx-\int sec^2x-1dx=\int 1dx-\int sec^2xdx-\int 1dx\]For the last line of the post above, it should be \(\large tan^2x=sec^2x-1\), sorry about that mistake

Answer 13

ok thanks that makes sense

Answer 14

np

Answer 15

Need a parantheses here\[\large \int 1dx-(\int sec^2(x)dx-\int1dx)\]So it would be\[\large \int 1dx-\int sec^2(x)dx+\int 1 dx=-tan(x)+2x+C\]