Def. and Indef. Trigonometric Integrals

• silicon_hobo
In summary, the student is attempting to solve the integralIn summary, the student is attempting to solve the integral\frac{1}{3}\sin^3\!x - \frac{1}{5}\sin^5\!x\^{\frac{\pi}{6}}_0\int^{\frac{\pi}{6}}_0 sin^2(x)\ cos^3 (x)\ dx\int tan^3(x)\ sec^4 (x)\ dxHomework Equations- The first one is solved using the same substitution method as in the previous post, and does not include the constant pi.- The second integral looks okay, but could
silicon_hobo

Homework Statement

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.
$$\int^{\frac{\pi}{6}}_0 sin^2(x)\ cos^3 (x)\ dx$$

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

The Attempt at a Solution

http://www.mcp-server.com/~lush/shillmud/int2.3a.JPG

http://www.mcp-server.com/~lush/shillmud/int2.3b.JPG

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:

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

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

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

silicon_hobo said:
Alright, I should already know this but what do you mean by 'transform your limits of integration'?

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...

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

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

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.

1. What is the difference between definite and indefinite trigonometric integrals?

Definite trigonometric integrals have specific limits of integration, while indefinite trigonometric integrals do not. This means that definite integrals have a numerical value, while indefinite integrals result in a function.

2. How do you solve a definite trigonometric integral?

To solve a definite trigonometric integral, you must first evaluate the indefinite integral and then substitute the limits of integration into the resulting function.

3. Are there any special rules or formulas for solving trigonometric integrals?

Yes, there are several special rules and formulas for solving trigonometric integrals, such as the power reduction formula, the half-angle formula, and the double angle formula.

4. Can trigonometric integrals be solved using substitution?

Yes, substitution is a commonly used method for solving trigonometric integrals. It involves replacing an expression within the integral with a new variable, making it easier to solve.

5. What are some real-world applications of trigonometric integrals?

Trigonometric integrals have many practical applications in fields such as physics, engineering, and astronomy. They can be used to calculate areas, volumes, and other physical quantities in real-world scenarios.

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