Integrate 1/cos(x): Wolfram Alpha Guide

[soopo]

How can you intgertate the 1/cos(x)?

The right answer can be found in Wolfram alpha at http://www.wolframalpha.com/input/?i=1/cos(x)

My first wrong answer was ln(cos(x)).
It suggests me that you cannot use the rule, ln(x), for trigonometric functions

 

[Bohrok]

There are several ways to integrate 1/cosx, or secx; just look on Google. You can try
\frac{1}{cosx}*\frac{cosx}{cosx} = \frac{cosx}{cos^2x} = \frac{cosx}{1-sin^2x}
and use u = sinx. Or working with secx, use the "clever substitution," as my calc book says, u = secx + tanx, du = sec²x + secxtanx dx. Then substitute u into the previous equation, get u and du together, and integrate.

 

[rzaidan]

Hi soopo
you can integrata as follows:
sec x + tan x
∫ (1/cos x) dx=∫ sec x dx=∫ sec x ______________ dx
sec x + tan x

(sec x)^2 + sec x tan x
∫ ___________________ dx
sec x + tan x
= ln(sec x + tan x) + C since the numerator is the derivative of the denominator.
Best Regards
Riad Zaidan

 

[soopo]

There are several ways to integrate 1/cosx, or secx; just look on Google. You can try
\frac{1}{cosx}*\frac{cosx}{cosx} = \frac{cosx}{cos^2x} = \frac{cosx}{1-sin^2x}
and use u = sinx. Or working with secx, use the "clever substitution," as my calc book says, u = secx + tanx, du = sec²x + secxtanx dx. Then substitute u into the previous equation, get u and du together, and integrate.

Thanks Bohrok!

I use this

cosx / (1 - (sinx)^2)

I get

1 / (1-u) du = ln|1-u|

Then, putting u=sinx back to the equation

1 / (1 - sinx) + C

---

This answer seems to differ from the answer in Wolfram Alpha.

 

[HallsofIvy]

Thanks Bohrok!

I use this

cosx / (1 - (sinx)^2)

I get

1 / (1-u) du = ln|1-u|

No, you get 1/(1- u^2) du

Then, putting u=sinx back to the equation

1 / (1 - sinx) + C

---

This answer seems to differ from the answer in Wolfram Alpha.

 

[Preno]

This may be worth remembering - the general substitution for integrating a rational function of sin, cos R(sin(x),cos(x)), which always works, is:

- t=cos(x) if R(-u,v)=-R(u,v)
- t=sin(x) if R(u,-v)=-R(u,v)
- t=tg(x) if R(-u,-v)=R(u,v)
- t=tg(x/2) in general

 

[soopo]

No, you get 1/(1- u^2) du

Thanks for the correction.

I get

I [ 1 / (1 - u^2) = .5 ln (1+sinx) - .5 ln (1 - sinx) + C

 

[soopo]

This may be worth remembering - the general substitution for integrating a rational function of sin, cos R(sin(x),cos(x)), which always works, is:

- t=cos(x) if R(-u,v)=-R(u,v)
- t=sin(x) if R(u,-v)=-R(u,v)
- t=tg(x) if R(-u,-v)=R(u,v)
- t=tg(x/2) in general

1. What is tg?

2. What is R(-u, v) = -R(u,v)?

 

[Preno]

tg = tan

R(u,v) is the rational function into which you plug sin(x) and cos(x) respectively. If R is "odd with respect to sin", you substitute for cos, and vice versa. t=tan(x/2) is the general substitution which always works (but can be rather cumbersome).

 

[drizzle]

edit, never mind :)