Integration using trigonometric identities

AmySaunders

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If \(\displaystyle \displaystyle{ \int_0^{\dfrac{\pi}{4}} } \, \tan^6(x)\, \sec(x)\, dx\, =\, I,\) express the value of \(\displaystyle \displaystyle{ \int_0^{\dfrac{\pi}{4}} } \, \tan^8(x)\, \sec(x)\, dx\) in terms of \(\displaystyle I.\)

I'm so lost with this problem. I don't even know where to begin.

Originally I thought I would find the value of the first equation and set it equal to I. Then I thought I could find the value of the second equation and divide by the value that is I to find the answer.

I don't know if this method is correct, and even if it is, I don't know how to find the value of the first equation.

Thank you for your help.
 
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If \(\displaystyle \displaystyle{ \int_0^{\dfrac{\pi}{4}} } \, \tan^6(x)\, \sec(x)\, dx\, =\, I,\) express the value of \(\displaystyle \displaystyle{ \int_0^{\dfrac{\pi}{4}} } \, \tan^8(x)\, \sec(x)\, dx\) in terms of \(\displaystyle I.\)

I'm so lost with this problem. I don't even know where to begin.

Originally I thought I would find the value of the first equation and set it equal to I. Then I thought I could find the value of the second equation and divide by the value that is I to find the answer.

I don't know if this method is correct, and even if it is, I don't know how to find the value of the first equation.

Thank you for your help.

tan8(x) = tan6(x)*[sec2(x) - 1]
 
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tan8(x) = tan6(x)*[sec2(x) - 1]

So then my second integral would be \(\displaystyle \displaystyle{ \int_0^{\dfrac{\pi}{4}} } \, \tan^6(x)\, \sec(x)\, \left(\sec^2(x)\, -\, 1\right)\, dx\)

What would my next step be?
 
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