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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 ?

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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 5 '14 at 17:42
The substitution $t=tan(\frac{x}{2})$ should also work. – Peter Feb 5 '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 5 '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:… – Michael Hardy Feb 5 '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. – Was Fr Feb 5 '14 at 18:17
up vote 3 down vote accepted

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 $$=\int (1+u^2)^3du \\=\int (1+3u^2+3u^4+u^6)du$$

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yes so I have now : $\int_{}^{}\frac{1}{(1+t^2)^3}dt$ how can I continue ? – Was Fr Feb 5 '14 at 17:53
Did you understand my first step – Sameh Shenawy Feb 5 '14 at 17:58
yes, I did my friend now I'm trying to solve the second step but still not moving ... – Was Fr Feb 5 '14 at 18:00
What about the rest – Sameh Shenawy Feb 5 '14 at 18:01
it seems connected some how to arctan derivative am I right ? – Was Fr Feb 5 '14 at 18:03

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