Question

Evaluate the following integrals: $$ \int \frac{d x}{\left(x^{2}-2 x \cos \theta+1\right)} $$

Short answer

Question: Evaluate the following integral, where θ is a constant: $$ \int \frac{dx}{x^2 - 2x\cos\theta + 1} $$ Answer: The evaluated integral is: $$ \int \frac{dx}{\left(x^{2}-2 x \cos \theta+1\right)} = \frac{1}{\sin\theta} \arctan\frac{x-\cos\theta}{\sin\theta} + C $$

Step-by-step solution

Observe the quadratic function

First, observe the given quadratic function in the denominator: $$ x^2 - 2x\cos\theta + 1 $$

Complete the square

In order to perform the integration, we need to complete the square in the denominator. Completing the square for the given quadratic function, we get: $$ \left(x- \cos\theta\right)^2 + (1-\cos^2\theta) $$

Trigonometric identity

Notice that we have the term \((1-\cos^2\theta)\) which can be simplified using the trigonometric identity \(\sin^2\theta+\cos^2\theta=1\). Thus we can rewrite the term as a \(\sin^2\theta\): $$ \left(x- \cos\theta\right)^2 + \sin^2\theta $$

Perform u-substitution

Now, perform the substitution \(u = x - \cos\theta\), then \(du = dx\). The integral can now be written as: $$ \int \frac{du}{u^2 + \sin^2\theta} $$

Apply the arctangent formula

We can now apply the arctangent formula for integration: $$ \int \frac{du}{u^2 + a^2} = \frac{1}{a} \arctan\frac{u}{a} + C $$ In the given integral, \(a = \sin\theta\). Thus, $$ \int \frac{du}{u^2 + \sin^2\theta} = \frac{1}{\sin\theta} \arctan\frac{u}{\sin\theta} + C $$

Substitute back for x

Finally, we substitute back the variable \(x\) using \(u = x - \cos\theta\): $$ \int \frac{dx}{x^2 - 2x\cos\theta + 1} = \frac{1}{\sin\theta} \arctan\frac{x-\cos\theta}{\sin\theta} + C $$ Therefore, the evaluated integral is: $$ \int \frac{dx}{\left(x^{2}-2 x \cos \theta+1\right)} = \frac{1}{\sin\theta} \arctan\frac{x-\cos\theta}{\sin\theta} + C $$