# Def. and Indef. Trigonometric Integrals

1. Mar 23, 2008

### silicon_hobo

1. The problem statement, all variables and given/known data
Hey folks, I believe I'm right in what I've done so far. I'm looking for confirmation of my methods and also wondering how to finish the first one by subbing in pi/6 for x and keeping pi in the answer. Thank you for your help. Cheers.
$$\int^{\frac{\pi}{6}}_0 sin^2(x)\ cos^3 (x)\ dx$$

$$\int tan^3(x)\ sec^4 (x)\ dx$$

2. Relevant equations

3. The attempt at a solution

2. Mar 23, 2008

### dynamicsolo

OK, on the first one, you did the same thing you did in the integral in your other post: don't forget to transform your limits of integration when you solve a definite integral by substitution!! (I bold-faced -- and double-exclamation-marked -- that because everybody makes that mistake at one time or another...)

The second integral looks okay.

Note, incidentally, that the second integral could also be busted apart as

(tan x)^2 · (sec x)^3 · sec x tan x dx ,

allowing the substitution u = sec x ;

writing (tan x)^2 as (sec x)^2 - 1 , you can also express the indefinite integral in powers of
sec x (one of many indefinite integrals that have more than one different-looking expression). IIRC, it is also one where, if you set out to show that the polynomial in tan x is the same as the polynomial in sec x, you get a left-over numerical constant... But since these are indefinite integrals and you have that arbitrary constant C hangin' around, the C obligingly "swallows up" the left-over number, so the two functions are effectively the same.

Last edited: Mar 23, 2008
3. Mar 23, 2008

### silicon_hobo

$$\frac{1}{3}\sin^3\!x - \frac{1}{5}\sin^5\!x\^{\frac{\pi}{6}}_0$$

. . , , $$= \;\bigg[\frac{1}{3}\left(\sin\frac{\pi}{6}\right)^3 - \frac{1}{5}\left(\sin\frac{\pi}{6}\right)^5\bigg] - \bigg[\frac{1}{3}\left(\sin0\right)^3 - \frac{1}{5}\left(\sin0\right)^5\bigg]$$

. . . . $$= \;\frac{1}{3}\left(\frac{1}{2}\right)^3 - \frac{1}{5}\left(\frac{1}{2}\right)^5 \;=\;\frac{1}{3}\!\cdot\!\frac{1}{8} - \frac{1}{5}\!\cdot\!\frac{1}{32} \;=\;\frac{17}{480}$$

4. Mar 23, 2008

### dynamicsolo

Sure, if you transform your indefinite integral back in terms of x first, you can use your original limits of integration x = 0 to x = pi/6 .

But when you shift over to u = sin x, your limits become

u = sin 0 = 0 to u = sin (pi/6) = 1/2 .

You have a nice theorem which says you can then just evaluate your indefinite integral

(u^3)/3 - (u^5)/5

from u = 0 to u = 1/2 , and get the correct result for the definite integral. (I agree with your answer 17/480.)

One thing I will warn you about is that if you're going to show the limits while you are working with u, you had better show the transformed limits with that integral, or a grader will certainly dock you for it...

5. Mar 23, 2008

### silicon_hobo

Okay, I've transformed the limits while using 'u' in the top image. Does that work?

6. Mar 23, 2008

### dynamicsolo

That looks better; a grader would be happy with either version. You could now go directly from the third line to the last to arrive at the value of the definite integral. The alternative is to solve the indefinite integral using the u-substitution, back-substitute sin x = u into the result to obtain the indefinite integral in terms of x, and then evaluate the definite integral using the limits in x. It is redundant to do both, so you could safely submit either version I've described.