Question

If $\int \frac{d\theta }{{\cos }^2\theta \left(\tan 2\theta +\sec 2\theta \right)}=\lambda \tan \theta +2{\log }_e\left|f\left(\theta \right)\right|+C$ , where $C$ is the constant of integration, then the ordered pair $\left(\lambda ,f\left(\theta \right)\right)$ is equal to:

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

Answer 1

The correct option is B

$\left(-1,1+\tan \theta \right)$

Explanation for the correct option:

Step 1: Simplify the left side of the given equation

The given integration equation is,

$\int \frac{d\theta }{{\cos }^2\theta \left(\tan 2\theta +\sec 2\theta \right)}=\lambda \tan \theta +2{\log }_e\left|f\left(\theta \right)\right|+C$

Let's consider, the left-hand side of the above equation, we get

$I=\int \frac{d\theta }{{\cos }^2\theta \left(\tan 2\theta +\sec 2\theta \right)}$

$\Rightarrow$ $I=\int \frac{{\sec }^2\theta d\theta }{\left({\displaystyle \frac{2\tan \theta }{1-{\tan }^2\theta }}+{\displaystyle \frac{1+{\tan }^2\theta }{1-{\tan }^2\theta }}\right)}$ $\left[\because {\sec }^2\theta =\frac{1}{{\cos }^2\theta },\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta },\sec 2\theta =\frac{1+{\tan }^2\theta }{1-{\tan }^2\theta }\right]$

$\Rightarrow$ $I=\int \frac{{\sec }^2\theta d\theta }{\left({\displaystyle \frac{{\tan }^2\theta +2\tan \theta +1}{1-{\tan }^2\theta }}\right)}$

$\Rightarrow$ $I=\int \frac{\left(1-{\tan }^2\theta \right){\sec }^2\theta d\theta }{{\left(1+\tan \theta \right)}^2}$

Step 2: Apply the substitute method

Substitute $\tan \theta =u$

$\Rightarrow$ ${\sec }^2\theta d\theta =du$

We get,

$I=\int \frac{\left(1-u^2\right)du}{{\left(1+u\right)}^2}$

$\Rightarrow$ $I=\int \frac{\left(1-u\right)du}{\left(1+u\right)}$

$\Rightarrow$ $I=\int \left(\frac{2}{1+u}-1\right)du$

$\Rightarrow$ $I=2\int \frac{1}{1+u}du-\int 1du$

$\Rightarrow$ $I=2{\log }_e\left|1+u\right|-u+C$

$\Rightarrow$ $I=2{\log }_e\left|1+\tan \theta \right|-\tan \theta +C$

Comparing the given equation, we get

$\lambda =-1$ and $f\left(\theta \right)=1+\tan \theta$

Thus, the value of the ordered pair $\left(\lambda ,f\left(\theta \right)\right)$ is $\left(-1,1+\tan \theta \right)$.

Hence, the correct option is $B)\left(-1,1+\tan \theta \right)$.