5
$\begingroup$

integral of $\int \theta\sec\theta \tan\theta \ d\theta$ enter image description here

my work

$\frac{d}{d\theta}\sec(θ) = \sec(\theta)\tan(\theta)$

So if we let $u = \theta$ and $v' = \sec(\theta)\tan(\theta)$, then we get:

$u = \theta, du = d\theta$ and $v = \sec(\theta), dv = \sec(\theta)\tan(\theta)d\theta$

Hence

$$\int \theta \sec(\theta)\tan(\theta) d\theta = \theta\sec(\theta) - \int\sec(\theta) d\theta $$

Now, the integral of $\sec(\theta)$ is a particularly tricky integral, but it comes to:

$$\int \sec(\theta) d \theta = \ln|\sec(\theta) + \tan(\theta)| + C$$

integral comes to:

$$\int \theta \sec(\theta)\tan(\theta) d\theta = \theta \sec(\theta) - \ln|\sec(\theta) + \tan(\theta)| + C $$

but my answer is not like this picture

enter image description here

please help me

$\endgroup$
  • $\begingroup$ Your work is fine. $\endgroup$ – user84413 Aug 26 '15 at 20:26
  • $\begingroup$ I need to be like the answer in the picture $\endgroup$ – Adel Hassan Aug 26 '15 at 20:27
  • 2
    $\begingroup$ @zorbha No, you don't need to: your answer is correct, the one in the book is not (because of missing absolute values). $\endgroup$ – egreg Aug 26 '15 at 20:47
7
$\begingroup$

Let $c = \cos \frac {\theta}2; s= \sin \frac {\theta}2$

Note that $$\sec \theta +\tan \theta = \frac {1+\sin \theta}{\cos \theta}=\frac {c^2+s^2+2cs}{c^2-s^2}=\frac {(c+s)^2}{(c+s)(c-s)}=\frac {c+s}{c-s}$$ and that should help you to reconcile the two answers.

$\endgroup$
  • $\begingroup$ @user84413 There were several typos! Thanks for pointing that out - I hope they are now sorted $\endgroup$ – Mark Bennet Aug 26 '15 at 20:31
5
$\begingroup$

Your answer is the same. Note that

$$\log|\cos(\theta/2)-\sin(\theta/2)|-\log|\cos(\theta/2)+\sin(\theta/2)|=\log |\frac{\cos(\theta/2)-\sin(\theta/2)}{\cos(\theta/2)+\sin(\theta/2)}|\\ =\log |\frac{\cos^2(\theta/2)-2\sin(\theta/2)\cos(\theta/2)+\sin^2(\theta/2)}{\cos^2(\theta/2)-\sin^2(\theta/2)}|\\=\log |\frac{1-\sin(\theta)}{\cos(\theta)}|=\log|\sec(\theta)-\tan(\theta)|= \log \frac{|\sec^2(\theta)-\tan^2(\theta)| }{|\sec(\theta)+\tan(\theta)|} \\=-\log| \sec(\theta)+\tan(\theta)|$$

$\endgroup$
0
$\begingroup$

An antiderivative of $$ \sec\theta\tan\theta=\frac{\sin\theta}{\cos^2\theta}= -\frac{-\sin\theta}{\cos^2\theta} $$ is $1/\cos\theta$. Therefore, integrating by parts, $$ \int\theta\frac{\sin\theta}{\cos^2\theta}\,d\theta= \frac{\theta}{\cos\theta}-\int\frac{1}{\cos\theta}\,d\theta $$ The remaining integral can be computed with the substitution $$ \theta=\frac{\pi}{2}-2u $$ so \begin{align} -\int\frac{1}{\cos\theta}\,d\theta= \int\frac{1}{\sin2u}\,2du &= \int\frac{\cos^2u+\sin^2u}{\sin u\cos u}\,du\\[6px] &= \int\left(\frac{\cos u}{\sin u}+\frac{\sin u}{\cos u}\right)du \\[6px] &= \log|\sin u|-\log|\cos u|+C\\[6px] &=\log\left|\tan u\right|+C\\[6px] &=\log\left|\tan\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right| \end{align} Now $$ \tan\left(\frac{\pi}{4}-\frac{\theta}{2}\right)= \frac{1-\tan(\theta/2)}{1+\tan(\theta/2)}= \frac{\cos(\theta/2)-\sin(\theta/2)}{\cos(\theta/2)+\sin(\theta/2)} $$ Thus the book is almost right: they're forgetting the absolute value.

You are stating that $$ \log|\sec\theta+\tan\theta| $$ is an antiderivative of $\sec\theta$; let's try: $$ \log|\sec\theta+\tan\theta|= \log\left|\frac{1+\sin\theta}{\cos\theta}\right|= \log|1+\sin\theta|-\log|\cos\theta| $$ The derivative is $$ \frac{\cos\theta}{1+\sin\theta}-\frac{-\sin\theta}{\cos\theta}= \frac{\cos^2\theta+\sin\theta+\sin^2\theta}{\cos\theta(1+\sin\theta)}= \frac{1+\sin\theta}{\cos\theta(1+\sin\theta)}=\frac{1}{\cos\theta} $$ So you're right as well. Even “more right” than the book.

$\endgroup$

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.