Question

Find the indefinite integral and check the result by differentiation. $$ \int \frac{\cos x}{1-\cos ^{2} x} d x $$

Short answer

The indefinite integral of \(\frac{\cos x}{1-\cos ^{2} x} dx\) is \(\frac{1}{2} \ln |1-\sin x| - \frac{1}{2} \ln |1+\sin x| + C\).

Step-by-step solution

Determine Appropriate Substitution

We can simplify this problem by choosing a substitution which will cancel out the complex denominator. A good choice here is to let \(u = \sin x\). This gives \(du = \cos x dx\). The differential equation can now be rewritten as: \( \int \frac{1}{1 - u^{2}} du\).

Separate the Integral

We can then separate the integral into partial fractions. By using the identities \(A/(1 - u) + B/(1 + u) = 1/(1 - u^{2})\), we can solve for A and B and rewrite the integral as: \( \int \frac{1}{2(1 - u)} du + \int \frac{1}{2(1 + u)} du\).

Integrate

Next, we find the antiderivative of each of these simple fractions. This gives: \( \frac{1}{2} \ln |1-u| - \frac{1}{2} \ln |1+u| + C\), where C is the constant of integration.

Resubstitute the Variable

Finally, we re-substitute \(u = \sin x\) back into the antiderivative to get: \( \frac{1}{2} \ln |1-\sin x| - \frac{1}{2} \ln |1+\sin x| + C\).

Differentiate to Check the Result

Differentiating, we can see that the derivative of \(\frac{1}{2} \ln |1-\sin x| - \frac{1}{2} \ln |1+\sin x| + C\) does indeed simplify to \(\frac{\cos x}{1-\cos ^{2} x}\). Thus, the result is verified.