Question

If, $\int \frac{d\theta}{cos^2\theta(tan2\theta+sec2\theta)} = \lambda tan\theta + 2log_e|f(\theta)| + c$ (where c is a constant of integration), then the ordered pair $(\lambda, |f(\theta)|)$ 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: (C)

Option C (-1, |1+tanθ|)

Answer 2

To solve the given integral, we'll follow these steps:

1. Substitution:

   Let $u = \tan\theta$. Then, $du = \sec^2\theta \, d\theta = \frac{d\theta}{\cos^2\theta}$.
   The integral becomes:

   $\int \frac{du}{\tan2\theta + \sec2\theta}$

2. Express $\tan2\theta$ and $\sec2\theta$ in terms of $u$:

   $\tan2\theta = \frac{2\tan\theta}{1-\tan^2\theta} = \frac{2u}{1-u^2}$

   $\sec2\theta = \frac{1}{\cos2\theta} = \frac{1+u^2}{1-u^2}$

   Then,

   $\tan2\theta + \sec2\theta = \frac{2u}{1-u^2} + \frac{1+u^2}{1-u^2} = \frac{2u+1+u^2}{1-u^2}$

3. Substitute back into the integral:

   $\int \frac{du}{\frac{2u+1+u^2}{1-u^2}} = \int \frac{1-u^2}{u^2+2u+1} \, du$

   Notice that $u^2+2u+1 = (u+1)^2$ and $1-u^2 = (1-u)(1+u)$. Thus,

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

4. Simplify and integrate:

   Write:

   $\frac{1-u}{1+u} = \frac{2}{1+u} - 1$

   Therefore,

   $\int \frac{1-u}{1+u} \, du = \int \left[ \frac{2}{1+u} - 1 \right] du = 2\int \frac{1}{1+u} \, du - \int 1 \, du$

   $\int \frac{1-u}{1+u} \, du = 2\ln|1+u| - u + C$

5. Substitute back $u = \tan\theta$:

   $-\tan\theta + 2\ln|1+\tan\theta| + C$

Comparing with the given form $\lambda \tan\theta + 2\ln|f(\theta)| + C$, we find:

$\lambda = -1, \quad f(\theta) = 1+\tan\theta$

Thus, the ordered pair $(\lambda, |f(\theta)|)$ is $(-1, |1+\tan\theta|)$.