Question

antiderivative of sec x(sec x+tan x)

Answer 1

\[\Large \int\limits_{ }^{ } \sec x (\sec x+ \tan x )~~dx\]\[\Large \int\limits_{ }^{ } \sec^2x + \tan x \sec x~~dx\]

Answer 2

we know that \[\frac{d}{dx}\sec(x)~=~\frac{d}{dx}(\cos x)^{-1}~=~-(\cos x)^{-2}(-\sin(x)=\]\[\sec^2x \sin x = \tan x \sec x\]

Answer 3

And we know that,\[\frac{d}{dx} \tan(x)=\frac{d}{dx} \frac{\sin(x)}{\cos(x)}=\frac{(\cos x)(\sin x) \prime - (\sin x)(\cos x) \prime}{\cos^2(x)}\]\[\frac{(\cos x)(\cos x)-(\sin x)(-\sin x)}{\cos^2x}=\frac{\cos^2x-(-\sin^2x)}{\cos^2x}=\]\[\frac{\cos^2x+\sin^2x}{\cos^2x}=\frac{1}{\cos^2x}=\sec^2x.\]

Answer 4

it takes long to show the derivaitves, but you get the point. Also don't forget the +C part.

Answer 5

reply if you have any questions to ask.

Answer 6

thanks a ton for the prompt reply..very helpful

Answer 7

you welcome !

Answer 8

so the antiderivative is sec2x+C?

Answer 9

no. not exactly.

Answer 10

we said tjhat the antiderivative of the sec^2x is tan(x)
and the antiderivative of tan(x)sec(x) is just sec(x).
So it comes,
\[\tan(x)+\sec(x)+C\]
But I think you can simplify that a bit to say,
\[\frac{\sin(x)}{\cos(x)}+\frac{1}{\cos(x)}+C\]\[\frac{\sin(x)+1}{\cos(x)}+C\]but it either way works.

Answer 11

oh ok