Introduction to the Integral Problem

We are given two definite integrals, \({I_1} = \mathop \smallint \limits_0^\pi \frac{{xdx}}{{1 + \sin x}}\) and \({I_2} = \mathop \smallint \limits_0^\pi \frac{{\left( {\pi - x} \right)dx}}{{1 - \sin \left( {\pi + x} \right)}}\). The question asks for the value of the integral \(I_1\).

Evaluating Integral \(I_1\)

Let's focus on evaluating the first integral, \(I_1\):

\({I_1} = \mathop \smallint \limits_0^\pi \frac{{x}}{{1 + \sin x}} dx\)

Using the Property of Definite Integrals

A common property used for evaluating definite integrals of this form, especially with limits from 0 to \(a\), is \(\mathop \smallint \limits_0^a f(x) dx = \mathop \smallint \limits_0^a f(a-x) dx\).

Here, the upper limit is \(\pi\). Let's apply this property to \(I_1\) by replacing \(x\) with \(\pi - x\):

\({I_1} = \mathop \smallint \limits_0^\pi \frac{{\left( {\pi - x} \right)}}{{1 + \sin \left( {\pi - x} \right)}} dx\)

We know from trigonometry that \(\sin \left( {\pi - x} \right) = \sin x\). Substituting this into the integral:

\({I_1} = \mathop \smallint \limits_0^\pi \frac{{\pi - x}}{{1 + \sin x}} dx\)

Combining the Integrals

Now we have two expressions for \(I_1\):

1. \({I_1} = \mathop \smallint \limits_0^\pi \frac{{x}}{{1 + \sin x}} dx\)
2. \({I_1} = \mathop \smallint \limits_0^\pi \frac{{\pi - x}}{{1 + \sin x}} dx\)

Adding these two expressions gives:

\(2{I_1} = \mathop \smallint \limits_0^\pi \frac{{x}}{{1 + \sin x}} dx + \mathop \smallint \limits_0^\pi \frac{{\pi - x}}{{1 + \sin x}} dx\)

Since the limits of integration and the denominator are the same, we can combine the numerators:

\(2{I_1} = \mathop \smallint \limits_0^\pi \frac{{x + \left( {\pi - x} \right)}}{{1 + \sin x}} dx\)

Simplifying the numerator:

\(2{I_1} = \mathop \smallint \limits_0^\pi \frac{{\pi}}{{1 + \sin x}} dx\)

We can take the constant \(\pi\) outside the integral:

\(2{I_1} = \pi \mathop \smallint \limits_0^\pi \frac{{1}}{{1 + \sin x}} dx\)

Evaluating the Simplified Integral

Now we need to evaluate the integral \(\mathop \smallint \limits_0^\pi \frac{{1}}{{1 + \sin x}} dx\). To do this, we can multiply the numerator and denominator by the conjugate of the denominator, which is \(1 - \sin x\):

\(\frac{1}{1 + \sin x} = \frac{1}{1 + \sin x} \times \frac{1 - \sin x}{1 - \sin x} = \frac{1 - \sin x}{1 - \sin^2 x}\)

Using the identity \(\sin^2 x + \cos^2 x = 1\), we have \(1 - \sin^2 x = \cos^2 x\):

\(\frac{1 - \sin x}{1 - \sin^2 x} = \frac{1 - \sin x}{\cos^2 x}\)

Now, we can split the fraction into two terms:

\(\frac{1}{\cos^2 x} - \frac{\sin x}{\cos^2 x} = \sec^2 x - \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} = \sec^2 x - \tan x \sec x\)

So, the integral becomes:

\(2{I_1} = \pi \mathop \smallint \limits_0^\pi \left( {\sec^2 x - \sec x \tan x} \right) dx\)

We know the standard integrals:

• \(\mathop \smallint \sec^2 x dx = \tan x + C\)
• \(\mathop \smallint \sec x \tan x dx = \sec x + C\)

Therefore, integrating term by term:

\(2{I_1} = \pi \left[ {\tan x - \sec x} \right]_0^\pi\)

Applying the Limits of Integration

Now we substitute the upper and lower limits of integration:

\(2{I_1} = \pi \left[ \left( {\tan \pi - \sec \pi} \right) - \left( {\tan 0 - \sec 0} \right) \right]\)

Evaluate the trigonometric functions at the limits:

• \(\tan \pi = 0\)
• \(\sec \pi = \frac{1}{\cos \pi} = \frac{1}{-1} = -1\)
• \(\tan 0 = 0\)
• \(\sec 0 = \frac{1}{\cos 0} = \frac{1}{1} = 1\)

Substitute these values back into the expression:

\(2{I_1} = \pi \left[ \left( {0 - \left( {-1} \right)} \right) - \left( {0 - 1} \right) \right]\)

\(2{I_1} = \pi \left[ \left( {0 + 1} \right) - \left( {-1} \right) \right]\)

\(2{I_1} = \pi \left[ {1 - \left( {-1} \right)} \right]\)

\(2{I_1} = \pi \left[ {1 + 1} \right]\)

\(2{I_1} = \pi \left[ {2} \right]\)

\(2{I_1} = 2\pi\)

Final Value of \(I_1\)

Finally, we solve for \(I_1\):

\(I_1 = \frac{2\pi}{2}\)

\(I_1 = \pi\)

Summary of Integral Evaluation Steps

• We started with the integral \(I_1 = \mathop \smallint \limits_0^\pi \frac{{x}}{{1 + \sin x}} dx\).
• We used the property \(\mathop \smallint \limits_0^a f(x) dx = \mathop \smallint \limits_0^a f(a-x) dx\) to get a second form of \(I_1\).
• We added the original and transformed integrals to simplify the numerator.
• We evaluated the simplified integral by multiplying the integrand by its conjugate and using trigonometric identities.
• We applied the limits of integration from 0 to \(\pi\).
• The final calculation yielded the value of \(I_1\).

Comparison with Provided Options

The calculated value for \(I_1\) is \(\pi\). Let's compare this with the given options:

• Option 1: 0
• Option 2: \(\frac{\pi}{2}\)
• Option 3: \(\pi\)
• Option 4: \(2\pi\)

Our calculated value matches Option 3.

Revision Table: Key Formulas for Integral Evaluation

Additional Information: Definite Integral Properties and Evaluation

Definite integrals represent the net signed area under a curve between two points. Evaluating them often involves finding the antiderivative of the function and applying the Fundamental Theorem of Calculus.

The property \(\mathop \smallint \limits_0^a f(x) dx = \mathop \smallint \limits_0^a f(a-x) dx\) is particularly useful when the integrand involves terms like \(x\) and functions of \(x\) that simplify nicely when \(x\) is replaced by \(a-x\), as seen in this problem where \(x + (\pi - x)\) simplifies to \(\pi\).

For integrals involving trigonometric functions in the denominator, multiplying by the conjugate (like \(1 - \sin x\) or \(1 - \cos x\)) is a common technique to transform the integrand into a form that is easier to integrate using standard formulas.