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how can I proceed with this exercise?

If

$$\int_{0}^{\frac{\pi}{4}} \tan^6(x) \sec(x) dx = I$$

then express

$$\int_{0}^{\frac{\pi}{4}} \tan^8(x) \sec(x) dx$$

in terms of $I$.

What I've got so far:

$$\int_{0}^{\frac{\pi}{4}} \tan^8(x) \sec(x) dx = \int_{0}^{\frac{\pi}{4}} \tan^2(x) \tan^6(x) \sec(x) dx = \int_{0}^{\frac{\pi}{4}} \left( \sec^2(x) - 1 \right) \tan^6(x) \sec(x) dx = \\ = - \int_{0}^{\frac{\pi}{4}} \tan^6(x) \sec(x) dx + \int_{0}^{\frac{\pi}{4}} \tan^6(x) \sec^3(x) dx = -I + \cdots$$

Any help is highly appreciated.

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(Exercise 50 from Stewart's Calculus book section chapter 7.2 7th edition)

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  • $\begingroup$ i would substitute $x=\arctan(y)$ and go by parts $\endgroup$ – tired Nov 6 '15 at 12:41
  • $\begingroup$ $I_6=\dfrac{13\sqrt2+15\ln\big(\sqrt2-1\big)}{48}~$ and $~I_8=-\dfrac{86\sqrt2+210\ln\big(\sqrt2-1\big)}{768}$ $\endgroup$ – Lucian Nov 6 '15 at 19:20
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Short answer: intergration by parts.

Long answer: denote the integral you want to find as $J$. Then you have

$J = \int_0^{\frac{\pi}{4}}\tan^8(x)\sec(x)dx = \int_0^{\frac{\pi}{4}}\tan^7(x)d(\sec(x)) = \left.\tan^7(x)\sec(x)\right|_0^{\frac{\pi}{4}} - \int_0^{\frac{\pi}{4}}\sec(x)d(\tan^7(x)) = \sqrt{2} - 7\int_0^{\frac{\pi}{4}}\tan^6(x)\sec^3(x)dx = \sqrt{2}-7\int_0^{\frac{\pi}{4}}\tan^6(x)\sec(x)(1+\tan^2(x))dx = \sqrt{2} - 7I - 7J = J$

$\sqrt{2} - 7I - 7J = J$

Hope all the steps are clear.

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Let's be a little nit more general and look at the Integrals

$$ I_{m}=\int \tan^m(x)\sec(x)dx\\ I_{m-2}=\int \tan^{m-2}(x)\sec(x)dx $$

now substitute $x=\arctan(y)$ . We get

$$ I_m=\int\frac{y^m}{\sqrt{1+y^2}}dy=\int y^{m-1}\frac{y}{\sqrt{1+y^2}}dy $$

Integrating by parts:

$$ I_m=y^{m-1}\sqrt{1+y^2}-(m-1)\int y^{m-2}\sqrt{1+y^2}dy=\\ y^{m-1}\sqrt{1+y^2}-(m-1)\int y^{m-2}\frac{1+y^2}{\sqrt{1+y^2}}dy=\\ y^{m-1}\sqrt{1+y^2}-(m-1)(I_m+I_{m-2}) $$

Therefore

$$ I_m=\frac{1}{m}y^{m-1}\sqrt{1+y^2}-\frac{m-1}{m}I_{m-2} $$

putting $m=8$, and plugging in the appropriate endpoints of integration $y=0$ and $y=1$ u can find your question as a special case of the above.

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  • $\begingroup$ definitely +1 a good answer. I have seen that question a while ago in that textbook and it got me thinking for some time... $\endgroup$ – imranfat Nov 7 '15 at 3:58

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