4
$\begingroup$

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?

$\endgroup$

3 Answers 3

5
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ 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, 2015 at 3:38
  • $\begingroup$ Yup, it's the chain rule. $\sec(11x)\tan(11x)$ is a single function of $11x$. $\endgroup$
    – user223391
    Apr 27, 2015 at 4:16
  • $\begingroup$ What if it was: sec(11x) tan(12x) ? $\endgroup$
    – dramadeur
    Apr 27, 2015 at 4:18
  • 1
    $\begingroup$ That's not the case here. In doing the chain rule, you treat $11x$ as a single variable. $\endgroup$
    – user223391
    Apr 27, 2015 at 4:19
3
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.