# if

Question:

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

1. (1) $(1,1-\tan \theta)$

2. (2) $(-1,1-\tan \theta)$

3. (3) $(-1,1+\tan \theta)$

4. (4) $(1,1+\tan \theta)$

Correct Option: , 3

Solution:

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

$=\int \frac{\sec ^{2} \theta}{\frac{1+\tan ^{2} \theta}{1-\tan ^{2} \theta}+\frac{2 \tan \theta}{1-\tan ^{2} \theta}} d \theta$

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

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

Let $\tan \theta=t \Rightarrow \sec ^{2} \theta d \theta=d t$, then

$I=\int\left(\frac{1-t}{1+t}\right) d t=\int\left(-1+\frac{2}{1+t}\right) d t$

$=-t+2 \log (1+t)+C$

$=-\tan \theta+2 \log (1+\tan \theta)+C$

Hence, by comparison $\lambda=-1$ and $f(x)=1+\tan \theta$