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I'm trying to integrate this function and I know that $\cos(x)$ in denominator with even power we should use z=tanx to solve the integral. But I'm not succeeding in solving it. Any ideas ?

$$\int\frac{1}{\cos^8 x} \ \mathrm{d}x$$

after reading the comments I came to that using t=tanx leads me to this integral :

$$\color{red}{\int (1+t^2)^3 dt}$$

how can we continue ?

Semsem
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Waseem Francis
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    Write the integrand as $\sec^2 x\sec^6 x$. Then write $\sec^6 x=(\sec^2 x)^3$ in terms of $\tan x$ using a Pythagorean identity. – David Mitra Feb 05 '14 at 17:42
  • The substitution $t=tan(\frac{x}{2})$ should also work. – Peter Feb 05 '14 at 17:47
  • Your final answer should come out to be $(\sin x/35\cos^7 x) (16\cos^6 x + 8 \cos^4 x +6 \cos^2 x + 5)$ – JPi Feb 05 '14 at 17:48
  • I edited the question by adding $dx$ in the title. If you're confused about why that matters, look at this: http://math.stackexchange.com/questions/200393/what-is-dx-in-integration/200403#200403 – Michael Hardy Feb 05 '14 at 18:08
  • I used $t=tan(\frac{x}{2})$ it just got harder so I see a nice answer down that I used thank you anyway. Michael thanks for pointing this out for me. – Waseem Francis Feb 05 '14 at 18:17

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you simply have $$\int \sec ^6x \sec^2 xdx=\int (1+\tan^2x)^3.\sec^2xdx$$ then use $u=\tan x$ and $du=\sec ^2xdx$ then $$=\int (1+u^2)^3du \\=\int (1+3u^2+3u^4+u^6)du$$

Semsem
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