Question

$\int \frac{dx}{\cot^2 x - 1} = \frac{1}{A} \log |\sec 2x + \tan 2x| - \frac{x}{B} + c$, (where $c$ is constant of integration), then $A + B =$}

• A. -6
• B. 6
• C. -5
• D. 5

Hint:

When integrating trigonometric expressions, always look for opportunities to simplify the integrand using fundamental identities, especially double-angle formulas ($\cos 2x$, $\sin^2 x$, $\cos^2 x$) to transform it into a more manageable form.

Answer 1

The Correct Option is A

Step 1: Simplify the integrand $\frac{1}{\cot^2 x - 1}$. Recall that $\cot^2 x - 1 = \frac{\cos^2 x}{\sin^2 x} - 1 = \frac{\cos^2 x - \sin^2 x}{\sin^2 x}$. Also, $\cos^2 x - \sin^2 x = \cos 2x$. So, $\frac{1}{\cot^2 x - 1} = \frac{1}{\frac{\cos 2x}{\sin^2 x = \frac{\sin^2 x}{\cos 2x}$.
Step 2: Use the identity $\sin^2 x = \frac{1 - \cos 2x}{2}$. \[ \int \frac{\sin^2 x}{\cos 2x} \,dx = \int \frac{\frac{1 - \cos 2x}{2{\cos 2x} \,dx = \frac{1}{2} \int \left(\frac{1}{\cos 2x} - \frac{\cos 2x}{\cos 2x}\right) \,dx \] \[ = \frac{1}{2} \int \left(\sec 2x - 1\right) \,dx \]
Step 3: Integrate term by term. Recall $\int \sec(ax) \,dx = \frac{1}{a} \log |\sec(ax) + \tan(ax)|$. \[ = \frac{1}{2} \left[\frac{1}{2} \log |\sec 2x + \tan 2x| - x\right] + c \] \[ = \frac{1}{4} \log |\sec 2x + \tan 2x| - \frac{x}{2} + c \]
Step 4: Compare the result with the given form $\frac{1}{A} \log |\sec 2x + \tan 2x| - \frac{x}{B} + c$. By comparison: \[ \frac{1}{A} = \frac{1}{4} \implies A = 4 \] \[ \frac{1}{B} = \frac{1}{2} \implies B = 2 \]
Step 5: Calculate $A + B$. \[ A + B = 4 + 2 = 6 \]