Definite trigonometric integral using properties

• Krushnaraj Pandya
In summary: I get it now!oh, right sorry! how silly of me. Thank you, that's a really good way to evaluate higher powers. I get it now!In summary, the conversation discusses finding the integral, ##I_{10}##, by using reduction formulae and properties of indefinite integrals. However, the process is tedious and the conversation also mentions using the binomial theorem to evaluate higher powers.

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Homework Statement

If ## I_n = \int_0^\frac {\pi}{4} \sec^n x dx## then find ## I_{10} - \frac {8}{9} I_8##

2. The attempt at a solution
this should be solvable by reduction formulae but since it'd be longer I wanted to know if there was a way to do it using mostly properties of indefinite integrals as suggested by my textbook. I'd appreciate the help, thank you

Krushnaraj Pandya said:

Homework Statement

If ## I_n = \int_0^\frac {\pi}{4} \sec^n x dx## then find ## I_{10} - \frac {8}{9} I_8##

2. The attempt at a solution
this should be solvable by reduction formulae but since it'd be longer I wanted to know if there was a way to do it using mostly properties of indefinite integrals as suggested by my textbook. I'd appreciate the help, thank you
I would start by finding ##I_2## and ##I_4## to see if I can deduce a pattern.

Mark44 said:
I would start by finding ##I_2## and ##I_4## to see if I can deduce a pattern.
I2 is tanx which evaluates to 1. I4 is tanx+(tanx^3) which is 4/3; I went further ahead to calculate I6 too, I guess the integration isn't very hard since you just have to substitute tanx as t everytime but I didn't see a pattern in the final-integrated functions

Krushnaraj Pandya said:
I2 is tanx which evaluates to 1. I4 is tanx+(tanx^3) which is 4/3; I went further ahead to calculate I6 too, I guess the integration isn't very hard since you just have to substitute tanx as t everytime but I didn't see a pattern in the final-integrated functions
I got the correct answer but the process was quite tedious

Krushnaraj Pandya said:
I got the correct answer but the process was quite tedious

Welcome to calculus (especially integration). Sometimes solutions are lengthy, and there is no way of shortening them. Of course, sometimes the opposite is true, IF some clever insights are used---not always, however.

Krushnaraj Pandya said:
you just have to substitute tanx as t everytime

If that's true, can you evaluate ##\int sec^{2k} x dx##?

(There's a Youtube video about developing a reduction formula for ##\int \sec^n x dx##. It writes ##\sec^n x## as ##(\sec^{n-2} x)( \sec^2 x)##. Then it uses integration by parts followed by using the identity ##\tan^2x = \sec^2x -1##. However, the presenter remarks "The integrals of even powers of secant are easy". So the person who posed your problem may have a simpler method in mind.)

Ray Vickson said:
Welcome to calculus (especially integration). Sometimes solutions are lengthy, and there is no way of shortening them. Of course, sometimes the opposite is true, IF some clever insights are used---not always, however.
hahaha, well I'm here to learn to get those intuitions at the end of the day I guess. (otherwise you could probably integrate anything by a long string of algorithmic substitutions etc)

Stephen Tashi said:
If that's true, can you evaluate ##\int sec^{2k} x dx##?

(There's a Youtube video about developing a reduction formula for ##\int \sec^n x dx##. It writes ##\sec^n x## as ##(\sec^{n-2} x)( \sec^2 x)##. Then it uses integration by parts followed by using the identity ##\tan^2x = \sec^2x -1##. However, the presenter remarks "The integrals of even powers of secant are easy". So the person who posed your problem may have a simpler method in mind.)
I can develop the reduction formula myself (but that'd still be lengthy if for example this question was in my entrance exams for a grad school- I'd never have the time to do it), but what I did was just write ##\sec^n x## as ##(\sec^{n-2} x)( \sec^2 x)## then directly put tanx=t instead of by parts (as done in the video). So I ended up having to solve ## (1 + t^2 )^4 ## for ##I_{10}## and then integrating the polynomial obtained. The same, just with power 3 for ##I_8##. There probably isn't anything quicker than this but I'm not sure (for example- what'd we do if we had sec^22, can't keep multiplying 1+t^2 20 times)

Use the binomial theorem.

vela said:
Use the binomial theorem.
I'm afraid that's two chapters ahead...so I only know the basics of binomial theorem till now- as in I can expand (1+x)^k etc., not sure how I can use it here

Krushnaraj Pandya said:
There probably isn't anything quicker than this but I'm not sure (for example- what'd we do if we had sec^22, can't keep multiplying 1+t^2 20 times)
Krushnaraj Pandya said:
I'm afraid that's two chapters ahead...so I only know the basics of binomial theorem till now- as in I can expand (1+x)^k etc., not sure how I can use it here
That's the form of ##(1+t^2)^{20}##, is it not?

vela said:
That's the form of ##(1+t^2)^{20}##, is it not?
oh, right sorry! how silly of me. Thank you, that's a really good way to evaluate higher powers.

1. What is a definite trigonometric integral?

A definite trigonometric integral is an integral that involves trigonometric functions, such as sine, cosine, or tangent, and has specific limits of integration. It is used to find the area under a curve or the value of a function at a given point.

2. What properties are used to solve definite trigonometric integrals?

The most commonly used properties for solving definite trigonometric integrals are the power rule, product rule, and quotient rule. These rules help to simplify the integral and make it easier to solve.

3. How do I use the power rule to solve a definite trigonometric integral?

The power rule states that the integral of x^n is (x^(n+1))/(n+1). To use this rule, you must first identify the power of the trigonometric function in the integral. Then, you can use the power rule to simplify the integral by raising the power by 1 and dividing by the new power.

4. Can I use substitution to solve a definite trigonometric integral?

Yes, substitution can be used to solve definite trigonometric integrals. This method involves substituting a new variable for the trigonometric function in the integral. This new variable should be chosen in a way that simplifies the integral and makes it easier to solve.

5. What is the difference between a definite and indefinite trigonometric integral?

The main difference between a definite and indefinite trigonometric integral is the presence of limits of integration. In a definite integral, the limits of integration are specified, whereas in an indefinite integral, the integration is performed without limits. This means that a definite integral will have a specific numerical value, while an indefinite integral will result in an equation with a constant term.