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How do you evaluate this trigonometric integral: $$\int 7\tan^5x\sec^2 x\,dx$$? Please help. Thank you in advance for your help.

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  • $\begingroup$ Put $u = \tan x$ $\endgroup$ – Sandeep Thilakan Feb 14 '14 at 5:39
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Given $\int 7\tan ^5 x \sec ^2 x dx$, let $\tan x=t$. Then $\sec ^2 x dx=dt$. Hence we must have $I=7\int t^5 dt= 7\frac{t^6}{6}+c=\frac{7}{6}\tan ^6 x+c$ where $c$ is constant.

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  • $\begingroup$ how did you know that tan(x) should equal t? Why not sec(x)? $\endgroup$ – rororo Feb 14 '14 at 5:51
  • $\begingroup$ Nice question. Well, note that the integrand is $\tan ^5 x\sec ^2 x dx$ (do not bother about 7, constant is not affected by integration so throw it out !). Now your substitution should be some thing so that its derivative will also be "adjusted". Here two trigonometric functions we can see , $\tan x$ and $\sec x$ with the powers 5 and 2 respectively. The derivative of $\tan x$ is $\sec ^2 x dx$ and THAT is already there. So if we choose $\tan x$ as $t$ then the whole integral will be converted to $\int t^5 dt$ which is standard integration. By choose $\sec x=t$ the result will be complicated $\endgroup$ – Anjan3 Feb 14 '14 at 5:59
  • $\begingroup$ I understand. Thank you! $\endgroup$ – rororo Feb 14 '14 at 6:02
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Since the exponent of $ \ \tan x \ $ is odd and that of $ \ \sec x \ $ is even, this "trigonometric powers integral" works the other way, too:

Choosing $ \ u \ = \ \sec x \ , $ we have $ \ du \ = \ \tan x \ \sec x \ dx \ $ and so may write

$$ 7 \int \ \tan^4 x \ \sec x \ \ (\tan x \ \sec x \ dx) \ = \ 7 \int \ (\tan^2 x )^2 \ \sec x \ \ (\tan x \ \sec x \ dx) $$

$$ 7 \int \ (\sec^2 x - 1)^2 \ \sec x \ \ (\tan x \ \sec x \ dx) \ \ \rightarrow \ \ 7 \int \ (u^2 -1 )^2 \ u \ \ du $$

$$ 7 \int \ u^5 \ - \ 2u^3 \ + \ u \ \ du \ = \ \frac{7}{6}u^6 \ - \ \frac{7}{2} u^4 \ + \ \frac{7}{2} u^2 \ + \ C $$

$$ \rightarrow \ 7 \ \sec^2 x \ (\frac{1}{6} \sec^4 x \ - \ \frac{1}{2} \sec^2 x \ + \ \frac{1}{2}) \ + \ C $$

[At this point, we have a perfectly acceptable "polynomial in secants"; but we should show that this in fact is equivalent to a simpler expression.]

$$ = \ 7 \ (\tan^2 x + 1) \ [ \ \frac{1}{6} (\tan^2 x + 1)^2 \ - \ \frac{1}{2} (\tan^2 x + 1) \ + \ \frac{1}{2} \ ] \ + \ C $$

$$ \ = \ 7 \ (\tan^2 x + 1) \ [ \ \frac{1}{6} \tan^4 x \ + \ \frac{1}{3} \tan^2 x \ + \ \frac{1}{6} \ - \ \frac{1}{2} \tan^2 x \ - \ \frac{1}{2} \ + \ \frac{1}{2} \ ] \ + \ C $$

$$ \ = \ \frac{7}{6} \ (\tan^2 x + 1) \ [ \ \tan^4 x \ - \ \tan^2 x \ \ + \ 1 \ ] \ + \ C $$

$$ \ = \ \frac{7}{6} \ ( \ \tan^6 x \ - \ \tan^4 x \ \ + \ \tan^2 x \ + \ \tan^4 x \ - \ \tan^2 x \ \ + \ 1 \ ) \ + \ C $$

$$ \ = \ \frac{7}{6} \tan^6 x \ + \ \frac{7}{6} \ + \ C \ = \ \frac{7}{6} \tan^6 x \ + \ C \ , $$

with the arbitrary constant "absorbing" the numerical one.

Now I grant that pretty much everybody would make the tangent-substitution instead. But it is worth mentioning that $ \ \int \ \tan^m x \ \sec^n x \ \ dx \ $ with $ \ m \ $ odd and $ \ n \ $ even can be computed with the secant-substitution as well. (One would probably only consider it, though, for small odd integer $ \ m \ $ .)

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Hint: Put $tan(x)=t$ and use integration by substitution

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$u$-substitute $u= \tan x$. Then, use power rule.

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