JEE PYQ: Indefinite Integration Question 18
Question 18 - 2020 (09 Jan Shift 2)
If $\int \frac{d\theta}{\cos^2\theta(\tan 2\theta + \sec 2\theta)} = \lambda\tan\theta + 2\log_e|f(\theta)| + C$ where $C$ is a constant of integration, then the ordered pair $(\lambda, f(\theta))$ is equal to:
(1) $(1, 1-\tan\theta)$
(2) $(-1, 1-\tan\theta)$
(3) $(-1, 1+\tan\theta)$
(4) $(1, 1+\tan\theta)$
Show Answer
Answer: (3)
Solution
$\tan 2\theta + \sec 2\theta = \frac{\sin 2\theta + 1}{\cos 2\theta} = \frac{(1+\tan\theta)^2}{1-\tan^2\theta} = \frac{1+\tan\theta}{1-\tan\theta}$. So $I = \int \frac{\sec^2\theta(1-\tan\theta)}{1+\tan\theta},d\theta$. Let $\tan\theta = t$: $\int \frac{1-t}{1+t},dt = \int\left(-1 + \frac{2}{1+t}\right),dt = -t + 2\ln|1+t| + C = -\tan\theta + 2\ln|1+\tan\theta| + C$. So $\lambda = -1$, $f(\theta) = 1+\tan\theta$.
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