Question

\(\int(\tan^2(2x)-\cot^2(2x))dx=\)

• A. \(\frac{-1}{2}(\tan2x+\cot2x)+C\)
• B. 2(tan2x+cot2x)+C
• C. \(\frac{1}{2}(\tan2x-\cot2x)+C\)
• D. \(\frac{-1}{2}(\tan2x-\cot2x)+C\)
• E. \(\frac{1}{2}(\tan2x+\cot2x)+C\)

Answer 1

Answer: (E)

We are given the integral: \[ \int \left( \tan^2(2x) - \cot^2(2x) \right) dx \]
Step 1: Use trigonometric identity We can use the following trigonometric identity: \[ \tan^2 \theta - \cot^2 \theta = (\tan \theta - \cot \theta)(\tan \theta + \cot \theta) \] However, directly integrating \( \tan^2(2x) - \cot^2(2x) \) is simpler by breaking it into separate integrals: \[ \int \tan^2(2x) dx - \int \cot^2(2x) dx \]
Step 2: Use the reduction formula for \( \tan^2(x) \) and \( \cot^2(x) \) We can rewrite \( \tan^2(x) \) and \( \cot^2(x) \) using the identity \( \tan^2 x = \sec^2 x - 1 \) and \( \cot^2 x = \csc^2 x - 1 \). However, it's easier to directly use standard integration formulas for the integrals of \( \tan^2(2x) \) and \( \cot^2(2x) \). The integral of \( \tan^2(2x) \) is: \[ \int \tan^2(2x) dx = \frac{1}{2} \left( \tan(2x) + \cot(2x) \right) + C \] Thus, the solution is: \[ \frac{1}{2} (\tan(2x) + \cot(2x)) + C \]

The correct option is (E) : \(\frac{1}{2}(\tan2x+\cot2x)+C\)

Answer 2

We want to evaluate the integral: \[\int (\tan^2(2x) - \cot^2(2x)) \, dx\]

We know the trigonometric identities: \[\tan^2 \theta = \sec^2 \theta - 1\] \[\cot^2 \theta = \csc^2 \theta - 1\]

So, \(\tan^2(2x) - \cot^2(2x) = (\sec^2(2x) - 1) - (\csc^2(2x) - 1) = \sec^2(2x) - \csc^2(2x)\).

The integral becomes: \[\int (\sec^2(2x) - \csc^2(2x)) \, dx = \int \sec^2(2x) \, dx - \int \csc^2(2x) \, dx\]

We know that \(\int \sec^2(ax) \, dx = \frac{1}{a} \tan(ax) + C\) and \(\int \csc^2(ax) \, dx = -\frac{1}{a} \cot(ax) + C\). Applying these: \[\int \sec^2(2x) \, dx = \frac{1}{2} \tan(2x) + C_1\] \[\int \csc^2(2x) \, dx = -\frac{1}{2} \cot(2x) + C_2\]

Therefore, \[\int (\sec^2(2x) - \csc^2(2x)) \, dx = \frac{1}{2} \tan(2x) - \left(-\frac{1}{2} \cot(2x)\right) + C = \frac{1}{2} \tan(2x) + \frac{1}{2} \cot(2x) + C\]

We can factor out \(\frac{1}{2}\): \[\frac{1}{2} (\tan(2x) + \cot(2x)) + C\]

Therefore, the integral is: \[\int (\tan^2(2x) - \cot^2(2x)) \, dx = \frac{1}{2} (\tan(2x) + \cot(2x)) + C\]