Trig intergration?

[Bobby Jones]

Using the substitution method and the quotient rule find the integral of:-


{integral sign} (1 + tan x) / (1 - tan x) dx

 

[galactus]

Use \(\displaystyle tan(x)=\frac{sin(x)}{cos(x)}\)

Simplify.

Then, use the sub \(\displaystyle u=cos(x)-sin(x)\)

It's easier than it looks. The solution involves ln.

 

[Bobby Jones]

Ok I used the quotient rule on sinx/cosx, in the numerator and denominator and got

(2- sinx^2)/cosx^2

Do I now intergrate this expression?

 

[galactus]

No.

There is no need to use the quotient rule on sin(x)/cos(x). That is a substitution.

\(\displaystyle \frac{1+\frac{sin(x)}{cos(x)}}{1-\frac{sin(x)}{cos(x)}}=\frac{cos(x)+sin(x)}{cos(x)-sin(x)}\)

Upon making the sub \(\displaystyle u=cos(x)-sin(x), \;\ -du=(cos(x)+sin(x))dx\)

The integral becomes:

\(\displaystyle \int\frac{1}{u}du\)

Integrate and resub.

 

[Bobby Jones]

Would this then become:

Ln(U)

Ln(Cosx - Sinx)


Is that the end, or could it go further?

 

[galactus]

That's it. Don't forget the negative sign.

You can leave it as is.

 

[Bobby Jones]

galactus,

How did you get the cosx + sinx to equal 1.

I know sin^2 x + cos^2 x = 1 , could you explain that move please.


so the answer would be

- Ln(Cosx + sinx) + C

 

[galactus]

I did not get cos(x)+sin(x)=1.

I just made the subs. That's all.

This problem is actually very easy. I don't think you're seeing it quite yet.

Let me start from the top.

\(\displaystyle \int\frac{1+tan(x)}{1-tan(x)}dx\)

Make the identity change \(\displaystyle tan(x)=\frac{sin(x)}{cos(x)}\)

\(\displaystyle \int\frac{1+\frac{sin(x)}{cos(x)}}{1-\frac{sin(x)}{cos(x)}}dx\)

Cross multiply the top and get \(\displaystyle 1+\frac{sin(x)}{cos(x)}=\frac{cos(x)+sin(x)}{cos(x)}\)

The bottom is then \(\displaystyle 1-\frac{sin(x)}{cos(x)}=\frac{cos(x)-sin(x)}{cos(x)}\)

So, we have:

\(\displaystyle \int\frac{\frac{cos(x)+sin(x)}{cos(x)}}{\frac{cos(x)-sin(x)}{cos(x)}}dx\)

\(\displaystyle \int\frac{cos(x)+sin(x)}{cos(x)}\cdot \frac{cos(x)}{cos(x)-sin(x)}dx\)

\(\displaystyle \int\frac{cos(x)+sin(x)}{cos(x)-sin(x)}dx\)

Now, make the sub \(\displaystyle u=cos(x)-sin(x), \;\ du=-(cos(x)+sin(x))dx\)

See?. The denominator is u and the numerator is the -du.

Your integral then becomes:

\(\displaystyle \int\frac{-du}{u}\) or \(\displaystyle -\int\frac{1}{u}du\)...however you want to write it.

Integrate:

\(\displaystyle -ln(u)+C\)

Resub:

\(\displaystyle -ln(cos(x)-sin(x))+C\)

That's it....

 

[lookagain]

Integrate: \(\displaystyle -ln(u)+C\) Resub: \(\displaystyle -ln(cos(x)-sin(x))+C\) That's it....

Everyone, be careful here.

\(\displaystyle This \ step \ must \ be \ -ln|u| + C, \ \ because \ the \ quantity \ that \ the \ logarithm\)
\(\displaystyle \ is \ being \ taken \ of \ could \ potentially \ be \ nonpositive.\)

\(\displaystyle \text{And, continuing, the answer would be:}\)

\(\displaystyle -ln|cos(x) - sin(x)| \ + \ C \ \ or\)

\(\displaystyle -ln|sin(x) - cos(x)| \ + \ C.\)



\(\displaystyle In \ comparison, \ for \ certain \ other \ problems, \ it \ would \ suffice \ to \ not\)
\(\displaystyle include \ the \ absolute \ value \ bars \ for \ those \ answers' \ final \ forms.\)

\(\displaystyle Example: \ \ \ ln(x^2 + 1) \ + \ C.\)

 

[Bobby Jones]

Cheers lads,

Whats the difference betwen the straight line bar and the bracket ( ) answer?

 

[lookagain]

Cheers lads,

Whats the difference betwen the straight line bar and the bracket ( ) answer?

\(\displaystyle The \ absolute \ value \ of \ x, \ |x|,\ equals \ \sqrt{x^2}.\)

On a number line, it is the (unsigned) distance from 0.

For examples:

|0| = 0

|-7| = 7

|3| = 3


\(\displaystyle Also, \ -|x| \ \ne \ |-x|.\)


However, with parentheses, brackets, and braces,

\(\displaystyle -(x) \ = \ -x, \ \ -[x] \ = \ [-x], \ \ -\{x\} \ = \ \{-x\}, \ respectively.\)