Integration by substitution - trig functions

[jonnburton]

Hi everyone,

I have a question about integrating trig functions in particular, but it might highlight my failure to properly understand the concept of integration in general.

The following is a worked example in my textbook:


Use the substitution s = sin x to find \(\displaystyle \int sin^4x cox^3x dx\)

\(\displaystyle \frac{ds}{dx}= cosx\) so \(\displaystyle ds = cosx dx\)

\(\displaystyle \int sin^4xcos^3x dx = \int sin^4xcos^2x(cosxdx)\)


This is where my question arises. Why does the \(\displaystyle cos^3x\) become \(\displaystyle cos^2x\)

I can see that there is a "cos x" at the end where the dx has been replaced by ds. But I can't see why this should affect the original function. (As I say, this is probably and indication of my failure to understand the concept of integration)...



Comparing that example with a simpler one:

\(\displaystyle \int(2x+7)^6\)

\(\displaystyle du = 2 dx\) so \(\displaystyle dx =\frac{1}{2}du\)

The integral becomes \(\displaystyle \frac{1}{2}\int u^6 du\) - the original function has not been altered...

Up until now I had thought that the dx was just a couple of letters tagged on to the end of an integral, just a formality (I'm sure this sounds flippant!) But now I'm starting to realise that it means something but I don't understand quite what..!

Is there any straightforward way of explaining this? Perhaps a online forum isn't the best place to ask this question as I don't know how complicated (or straightforward) this really is. But I would like to understand properly the concepts dealt with and so would be really grateful if anyone could shed some light on this!

 

[HallsofIvy]

Hi everyone,

I have a question about integrating trig functions in particular, but it might highlight my failure to properly understand the concept of integration in general.

The following is a worked example in my textbook:


Use the substitution s = sin x to find \(\displaystyle \int sin^4x cox^3x dx\)

\(\displaystyle \frac{ds}{dx}= cosx\) so \(\displaystyle ds = cosx dx\)

\(\displaystyle \int sin^4xcos^3x dx = \int sin^4xcos^2x(cosxdx)\)


This is where my question arises. Why does the \(\displaystyle cos^3x\) become \(\displaystyle cos^2x\)

I can see that there is a "cos x" at the end where the dx has been replaced by ds. But I can't see why this should affect the original function. (As I say, this is probably and indication of my failure to understand the concept of integration)...

No, it has nothing to do with integration. It has to do with algebra. For any x, \(\displaystyle x^3= x(x)(x)\), by definition.
\(\displaystyle cos^3(x)= (cos(x))(cos(x))(cos(x))= [(cos(x))(cos(x))](cos(x))= cos^2(x) cos(x)\).

Comparing that example with a simpler one:

\(\displaystyle \int(2x+7)^6\)

You mean \(\displaystyle \int(2x+7)^6 dx\). Never write an integral without its "dx".

\(\displaystyle du = 2 dx\) so \(\displaystyle dx =\frac{1}{2}du\)

The integral becomes \(\displaystyle \frac{1}{2}\int u^6 du\) - the original function has not been altered...

Yes, it has. It has been altered from 2x+ 7 to (1/2)u.

Up until now I had thought that the dx was just a couple of letters tagged on to the end of an integral, just a formality (I'm sure this sounds flippant!) But now I'm starting to realise that it means something but I don't understand quite what..!

Is there any straightforward way of explaining this? Perhaps a online forum isn't the best place to ask this question as I don't know how complicated (or straightforward) this really is. But I would like to understand properly the concepts dealt with and so would be really grateful if anyone could shed some light on this!

Have you seen the "Riemann sum" definition of an integral? To integrate f(x) from a to b, divide the x-axis into n subintervals. \(\displaystyle \Delta x\) is the length of each subinterval so, taking \(\displaystyle x_i\) to be some point in that subinterval \(\displaystyle f(x_i)\Delta x\) can be thought of a "base times height" so "area" of that rectangle. Adding them gives \(\displaystyle \sum f(x_i)\Delta x\), approximating the "area" under the graph. Taking the limit as \(\displaystyle \Delta x\) gets smaller and smaller, becoming "dx", gives the exact "area", the integral. In effect, \(\displaystyle \Delta x\) tells how we are measuring along the x-axis. When we to "u" instead of "x" we are measuring that differently.

If, for example, to integrate \(\displaystyle \int_0^3 (2x+7)^6 dx\), you use the substitution u= 2x+ 7, when x= 0, u= 2(0)+ 7= 7 and when x= 3, u= 2(3)+ 7= 13. The distance over which you are integrating is now 7 to 13= 6 rather than 0 to 3= 3. It has been doubled which you allow for by dividing the function by 2.

 

[jonnburton]

Thanks, HallsofIvy. I am going to spend a few days reviewing integration right from the start to get to grips with this, as I keep getting confused!