integral of secx

\displaystyle\ resolve

\displaystyle\int_0^{35^{\circ}} \ sec(x) \ dx \longrightarrow

\displaystyle\int_0^{35^{\circ}}\ sec(x) \frac{sec(x)+tan(x)}{sec(x)+tan(x)} \ dx

....obviously ( I was told this was the first step)

\displaystyle\ sec(x)+tan(x) = u

\displaystyle\frac{du}{dx} = sec(x)\times tan(x) + sec^2(x) \ and \ then \ I'm \ stuck

Reply 1

Farhan.Hanif93

Original post by wcp100

\displaystyle\ resolve

\displaystyle\int_0^{35^{\circ}} \ sec(x) \ dx \longrightarrow

\displaystyle\int_0^{35^{\circ}}\ sec(x) \frac{sec(x)+tan(x)}{sec(x)+tan(x)} \ dx

....obviously ( I was told this was the first step)

\displaystyle\ sec(x)+tan(x) = u

\displaystyle\frac{du}{dx} = sec(x)\times tan(x) + sec^2(x) \ and \ then \ I'm \ stuck

Immediately there's a problem as your limits do not make sense. They must either be in radians or as a fraction of 360.

What you need to note is that your integrand can be rewritten as:

\dfrac{\sec^2 x + \sec x \tan x}{\tan x + \sec x}

What is the derivative of

\tan x + \sec x

? Does it look familiar? What do you do with integrals with integrands of the form

\dfrac{f'(x)}{f(x)}

? *cough* natural logarithm *cough*.

(edited 13 years ago)

Reply 2

kfkle

Recall the integral of f'(x)/f(x).

Reply 3

a²+b² = c²

I don't like that way of finding the integral of the secant.
It assumes that we already know the answer, doesn't it?
Why the hell would we multiply by that otherwise? :dontknow:

Reply 4

nosignal456

this needs to be moved to the health and relationships forum.

Reply 5

bananarama2

Original post by a²+b² c²

I don't like that way of finding the integral of the secant.
It assumes that we already know the answer, doesn't it?
Why the hell would we multiply by that otherwise? :dontknow:

You were thinking exactly what I was. Is there another way of doing it ?

Reply 6

Farhan.Hanif93

Original post by a²+b² c²

I don't like that way of finding the integral of the secant.
It assumes that we already know the answer, doesn't it?
Why the hell would we multiply by that otherwise? :dontknow:

Original post by wcp100

You were thinking exactly what I was. Is there another way of doing it ?

Yes, there is.

Note that

\sec x = \dfrac{1}{\cos x} = \dfrac{\cos x}{1-\sin ^2x} = \dfrac{\cos x}{(1+\sin x)(1-\sin x)}

.

Then use a substitution of

u =\sin x

and use partial fractions.

Reply 7

roar558

Original post by wcp100

\displaystyle\ resolve

\displaystyle\int_0^{35^{\circ}} \ sec(x) \ dx \longrightarrow

\displaystyle\int_0^{35^{\circ}}\ sec(x) \frac{sec(x)+tan(x)}{sec(x)+tan(x)} \ dx

....obviously ( I was told this was the first step)

\displaystyle\ sec(x)+tan(x) = u

\displaystyle\frac{du}{dx} = sec(x)\times tan(x) + sec^2(x) \ and \ then \ I'm \ stuck

well you could just substitute in du for

sec(x)\times tan(x) + sec^2(x)

and u for

\displaystyle\ sec(x)+tan(x)

to get

\displaystyle\int_0^{35^{\circ}} \frac{1}{u} \ du

Reply 8

$Avatar for Get me off the £\?%!^@ computer$

Get me off the £\?%!^@ computer

Original post by wcp100

You were thinking exactly what I was. Is there another way of doing it ?

You can also use the well known t substitution.

t=tan(x/2)

cos x = (1-t^2)/(1+t^2)

dx/dt=2/(1+t^2)

It's all very straightforward and requires no cleverness at all.