If ∫ dθ ^ 2 θ( 2θ + 2θ) = λ tanθ + 2 _ e |f(θ)| + C where C is a...

$∫ \dfrac{\sec^{2}\theta}{\dfrac{1 + \tan^{2}\theta}{1 - \tan^{2}\theta} + \dfrac{2tan\theta}{1 - \tan^{2}\theta}}d\theta$

$= ∫ \dfrac{\sec^{2}\theta\left(1 - \tan^{2}\theta\right)}{{\left(1 + tan\theta\right)}^{2}}d\theta$

$= ∫ \dfrac{\sec^{2}\theta\left(1 - tan\theta\right)}{1 + tan\theta}d\theta$

Let $tan\theta = t ⇒ \sec^{2}\thetad\theta = dt$

$= ∫ \left(\dfrac{1 - t}{1 + t}\right)dt = ∫ \left(- 1 + \dfrac{2}{1 + t}\right)dt$

$= - t + 2\ln \left(1 + t\right) + C$

$= - tan\theta + 2\ln \left(1 + tan\theta\right) + C$

$⇒ \lambda = - 1andf\left(\theta\right) = 1 + tan\theta$