Step 1: Understanding the Concept:
We are tasked with evaluating a definite integral involving trigonometric functions.
This can be approached either by using standard properties of definite integrals or through a suitable substitution.
Step 2: Key Formula or Approach:
Method 1 relies on the property: \(\int_{0}^{2a} f(x) dx = 0\) if \(f(2a-x) = -f(x)\).
Method 2 utilizes substitution: Let \(t = \sin x\), which implies \(dt = \cos x dx\).
Step 3: Detailed Explanation:
Let the integral be \(I = \int_{0}^{\pi} \frac{\cos x}{1+\sin^2 x} dx\).
Method 1: Using Definite Integral Properties
Let the integrand be denoted as \(f(x) = \frac{\cos x}{1+\sin^2 x}\).
We evaluate the function at the upper limit minus \(x\), which is \(f(\pi - x)\):
\[ f(\pi - x) = \frac{\cos(\pi - x)}{1+\sin^2(\pi - x)} \]
Recall the trigonometric identities: \(\cos(\pi - x) = -\cos x\) and \(\sin(\pi - x) = \sin x\).
Substituting these identities back, we get:
\[ f(\pi - x) = \frac{-\cos x}{1+(\sin x)^2} = -\frac{\cos x}{1+\sin^2 x} = -f(x) \]
Since \(f(\pi - x) = -f(x)\), it directly follows from the properties of definite integrals that the integral over the interval \([0, \pi]\) is zero.
\[ \int_{0}^{\pi} f(x) dx = 0 \]
Method 2: Using Substitution
Let's employ the substitution \(t = \sin x\).
Differentiating both sides with respect to \(x\) gives \(dt = \cos x dx\).
Next, we must change the limits of integration to match the new variable \(t\).
When the lower limit \(x = 0\), \(t = \sin(0) = 0\).
When the upper limit \(x = \pi\), \(t = \sin(\pi) = 0\).
Substituting these into the original integral transforms it to:
\[ I = \int_{0}^{0} \frac{1}{1+t^2} dt \]
Because the upper and lower limits of integration are identical, the definite integral evaluates to \(0\).
Step 4: Final Answer:
The value of the integral is \(0\).