# Definite trigonometric integral using properties

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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

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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
I would start by finding ##I_2## and ##I_4## to see if I can deduce a pattern.

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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

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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

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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.

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.)

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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)

Gold Member
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, cant keep multiplying 1+t^2 20 times)

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Homework Helper
Use the binomial theorem.

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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

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