Question

If $\int \frac{d\theta}{cos^{2}\,\theta\left(tan\,2\theta+sec\,2\theta\right)}=$
$\lambda\,tan\theta+2\,log_{e}\left|f \left(\theta\right)+C\right|$ 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)

$I = \int \frac{d\theta}{cos^{2}\theta\left(tan\,2\theta+sec\,2\theta \right) }$
$= \int \frac{sec^{2}\,\theta \,d\theta}{\frac{2\,tan\,\theta}{1-,tan^{2}\,\theta }+\frac{1+,tan^{2}\,\theta }{1-,tan^{2}\,\theta } = \int \frac{\left(1-,tan^{2}\,\theta \right)sec^{2}\,\theta \,d\theta }{\left(1+tan\,\theta \right)^{2}}}$
$tan\,\theta = t \Rightarrow sec^{2}\,\theta d\,\theta = dt$
$I = \int\frac{1-t^{2}}{\left(1+t\right)^{2}}dt = \int \frac{\left(1-t\right)\left(1+t\right)}{\left(1+t\right)^{2}}dt$
$= \int \frac{1}{1+t}-\frac{1}{1+t}dt$
$= \ell n \left|1 +t\right|-\int \left(\frac{1+t}{1+t}-\frac{1}{1+t}\right)dt$
$ \ell n \left|1 +t\right|-t+ \ell n \left|1 +t\right|$
$ = \ell2 n \left|1 +tan\theta\right| - tan\theta +C$
$\lambda = -1, f\left(\theta\right)= 1 + tan\theta$