Solve the integral of cos(2x) / (1 - cos^2(2x)) dx.
Understand the Problem
The question is asking to solve the integral of cos(2x) / (1 - cos^2(2x)) with respect to x. We'll need to apply trigonometric identities to simplify and evaluate the integral.
Answer
$-\frac{1}{2} \csc{2x} + C$
Answer for screen readers
$$ -\frac{1}{2} \csc{2x} + C $$
Steps to Solve
- Apply the Pythagorean trigonometric identity
Recall the Pythagorean identity: $\sin^2{\theta} + \cos^2{\theta} = 1$. From this, we can deduce that $1 - \cos^2{\theta} = \sin^2{\theta}$. Apply this to our integral:
$$ \int \frac{\cos{2x}}{1 - \cos^2{2x}} dx = \int \frac{\cos{2x}}{\sin^2{2x}} dx $$
- Rewrite the integrand
Separate the fraction into two parts:
$$ \int \frac{\cos{2x}}{\sin^2{2x}} dx = \int \frac{1}{\sin{2x}} \cdot \frac{\cos{2x}}{\sin{2x}} dx = \int \csc{2x} \cot{2x} dx $$
- Perform u-substitution
Let $u = 2x$, then $du = 2 dx$, so $dx = \frac{1}{2} du$. $$ \int \csc{2x} \cot{2x} dx = \int \csc{u} \cot{u} \cdot \frac{1}{2} du = \frac{1}{2} \int \csc{u} \cot{u} du $$
- Evaluate the integral
The integral of $\csc{u} \cot{u}$ is $-\csc{u}$. Therefore:
$$ \frac{1}{2} \int \csc{u} \cot{u} du = \frac{1}{2} (-\csc{u}) + C = -\frac{1}{2} \csc{u} + C $$
- Substitute back
Substitute $u = 2x$ back into the expression:
$$ -\frac{1}{2} \csc{u} + C = -\frac{1}{2} \csc{2x} + C $$
$$ -\frac{1}{2} \csc{2x} + C $$
More Information
The final answer is $-\frac{1}{2} \csc{2x} + C$, where $C$ is the constant of integration.
Tips
A common mistake is forgetting the constant of integration, $C$. Another mistake is incorrectly applying the chain rule during the u-substitution step, especially when dealing with the $2x$ term. Also, students might incorrectly remember the integral of $ \csc{x} \cot{x} $.
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