integral of $\int \theta\sec\theta \tan\theta \ d\theta$
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
please help me