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Evaluate the indefinite integral:

$\displaystyle \int \frac{\sec(11 x) \tan(11 x)}{\sqrt{\sec(11 x)}} \, dx $

(using substitution)

The answer is: $\frac{2}{11} \sqrt{sec(11 x)} + C$

I don't get where $11$ in $\frac{2}{11}$ comes from

My solution:

u = sec($11x$)

du = sec(11x) $\cdot$ tan(11x) dx

Making substitution:

$\displaystyle \int \frac{1}{\sqrt{u}} du$

Evaluating integral:

$\frac {u^\frac{1}{2}}{\frac {1}{2}} -> 2 \cdot \sqrt {sec(11x)} + C$

As you can see, there shouldn't 11 in the answer... but how come there is?

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Your $du$ is wrong. It should have an extra 11 in it. :)

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  • $\begingroup$ Is it because of chain rule? I'm not sure how it's applied if there are several terms that are multiplied together $\endgroup$ – dramadeur Apr 27 '15 at 3:38
  • $\begingroup$ Yup, it's the chain rule. $\sec(11x)\tan(11x)$ is a single function of $11x$. $\endgroup$ – user223391 Apr 27 '15 at 4:16
  • $\begingroup$ What if it was: sec(11x) tan(12x) ? $\endgroup$ – dramadeur Apr 27 '15 at 4:18
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    $\begingroup$ That's not the case here. In doing the chain rule, you treat $11x$ as a single variable. $\endgroup$ – user223391 Apr 27 '15 at 4:19
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Just to make things easier on my eyes (I hate all of the 11's) i'd start off with:

$let\:v = 11x,\: dv = 11dx$

$$\frac{1}{11}\int{\frac{\sec v \:\tan v}{\sqrt{\sec v}}}\:dv$$

$let\: u=\sec v,\: du = \sec v\:\tan v\:dv$

$$\frac{1}{11}\int{\frac{1}{\sqrt{u}}}\:du$$

You can take it from here.

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$du = sec(11x) tan (11x) (11dx)$

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