[Math] Evaluate $\int \theta\sec\theta \tan\theta \ d\theta$

calculusintegration

integral of $\int \theta\sec\theta \tan\theta \ d\theta$
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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

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please help me

Best Answer

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.

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