Question

If \[ \int \frac{1+\tan x}{\sin 2x}\,dx=A\log \tan x+B\tan x+C, \] then \(4A-2B=\)

• A. \(-1\)
• B. \(2\)
• C. \(1\)
• D. \(-2\)

Hint:

For integrals involving \(\tan x\) and \(\sin 2x\), use \(\sin 2x=\dfrac{2\tan x}{1+\tan^2x}\) and substitute \(t=\tan x\).

Answer 1

The Correct Option is C

Step 1: Use the identity for \(\sin 2x\).
We know that \[ \sin 2x=\frac{2\tan x}{1+\tan^2 x} \] Let \[ t=\tan x \] Then, \[ dt=\sec^2x\,dx=(1+\tan^2x)\,dx \] So, \[ dt=(1+t^2)\,dx \] \[ dx=\frac{dt}{1+t^2} \]

Step 2: Substitute in the integral.
Given integral is \[ \int \frac{1+\tan x}{\sin 2x}\,dx \] Using \(t=\tan x\), \[ \int \frac{1+t}{\dfrac{2t}{1+t^2}}\cdot \frac{dt}{1+t^2} \] \[ = \int \frac{1+t}{2t}\,dt \] \[ = \frac{1}{2}\int \left(\frac{1}{t}+1\right)\,dt \]

Step 3: Integrate.
\[ \frac{1}{2}\int \left(\frac{1}{t}+1\right)\,dt = \frac{1}{2}\log t+\frac{t}{2}+C \] Substituting back \(t=\tan x\), \[ = \frac{1}{2}\log \tan x+\frac{1}{2}\tan x+C \]

Step 4: Compare with the given form.
Given, \[ \int \frac{1+\tan x}{\sin 2x}\,dx=A\log \tan x+B\tan x+C \] Comparing, \[ A=\frac{1}{2} \] and \[ B=\frac{1}{2} \]

Step 5: Find \(4A-2B\).
\[ 4A-2B=4\left(\frac{1}{2}\right)-2\left(\frac{1}{2}\right) \] \[ =2-1 \] \[ =1 \]

Step 6: Final conclusion.
Hence, \[ \boxed{1} \]