Question

If $\int \frac{d\theta }{co{s}^2\theta \left(tan2\theta +sec2\theta \right)}=$
$\lambda tan\theta +2lo{g}_e\left|f\left(\theta \right)+C\right|$ where C is a constant of integration, then the ordered pair $\left(\lambda ,f\left(\theta \right)\right)$ is equal to :

• A. $(-1,1-tan\theta )$
• B. $(-1,1+tan\theta )$
• C. $(1,1+tan\theta )$
• D. $(1,1-tan\theta )$

Answer 1

Answer: (B)

$I=\int \frac{d\theta }{co{s}^2\theta \left(tan2\theta +sec2\theta \right)}$
$=\int \frac{se{c}^2\theta d\theta }{\frac{2tan\theta }{1-,ta{n}^2\theta }+\frac{1+,ta{n}^2\theta }{1-,ta{n}^2\theta }=\int \frac{\left(1-,ta{n}^2\theta \right)se{c}^2\theta d\theta }{{\left(1+tan\theta \right)}^2}}$
$tan\theta =t\Rightarrow se{c}^2\theta d\theta =dt$
$I=\int \frac{1-t^2}{{\left(1+t\right)}^2}dt=\int \frac{\left(1-t\right)\left(1+t\right)}{{\left(1+t\right)}^2}dt$
$=\int \frac{1}{1+t}-\frac{1}{1+t}dt$
$=\ell n\left|1+t\right|-\int \left(\frac{1+t}{1+t}-\frac{1}{1+t}\right)dt$
$\ell n\left|1+t\right|-t+\ell n\left|1+t\right|$
$=\ell 2n\left|1+tan\theta \right|-tan\theta +C$
$\lambda =-1,f\left(\theta \right)=1+tan\theta$