Understanding the Given Integrals

The problem asks for the value of the sum of two definite integrals, \(I_1\) and \(I_2\). The integrals are given by:

• \(I_1 = \mathop \smallint \limits_0^\pi \frac{{xdx}}{{1 + \sin x}}\)
• \(I_2 = \mathop \smallint \limits_0^\pi \frac{{\left( {\pi - x} \right)dx}}{{1 - \sin \left( {\pi + x} \right)}}\)

We need to find the value of \(I_1 + I_2\).

Simplifying Integral I2

Before adding the integrals, let's simplify the integrand of \(I_2\). The denominator of \(I_2\) involves \(\sin(\pi + x)\). We can use the trigonometric identity for \(\sin(\pi + \theta)\).

Recall the identity: \(\sin(\pi + \theta) = -\sin \theta\).

Using this, we have \(\sin(\pi + x) = -\sin x\). Substituting this into the expression for \(I_2\):

\(I_2 = \mathop \smallint \limits_0^\pi \frac{{\left( {\pi - x} \right)dx}}{{1 - (-\sin x)}}\)

\(I_2 = \mathop \smallint \limits_0^\pi \frac{{\left( {\pi - x} \right)dx}}{{1 + \sin x}}\)

Now, both \(I_1\) and the simplified \(I_2\) have the same denominator, \(1 + \sin x\).

Calculating the Sum I1 + I2

Now we can add the two integrals:

\(I_1 + I_2 = \mathop \smallint \limits_0^\pi \frac{{xdx}}{{1 + \sin x}}\;\; + \mathop \smallint \limits_0^\pi \frac{{\left( {\pi - x} \right)dx}}{{1 + \sin x}}\)

Since the limits of integration (from 0 to \(\pi\)) and the denominators are the same, we can combine the integrands into a single integral:

\(I_1 + I_2 = \mathop \smallint \limits_0^\pi \frac{{x + (\pi - x)}}{{1 + \sin x}}dx\)

Simplify the numerator:

\(I_1 + I_2 = \mathop \smallint \limits_0^\pi \frac{{\pi}}{{1 + \sin x}}dx\)

We can factor out the constant \(\pi\) from the integral:

\(I_1 + I_2 = \pi \mathop \smallint \limits_0^\pi \frac{1}{{1 + \sin x}}dx\)

Evaluating the Definite Integral \( \mathop \smallint \limits_0^\pi \frac{1}{{1 + \sin x}}dx \)

To evaluate the definite integral \(\mathop \smallint \limits_0^\pi \frac{1}{{1 + \sin x}}dx\), we can use the substitution \(t = \tan(x/2)\). This is a common technique for integrals involving trigonometric functions.

If \(t = \tan(x/2)\), then:

• \(dx = \frac{2dt}{1+t^2}\)
• \(\sin x = \frac{2t}{1+t^2}\)

We also need to change the limits of integration:

• When \(x = 0\), \(t = \tan(0/2) = \tan(0) = 0\).
• When \(x = \pi\), \(t = \tan(\pi/2)\), which tends to infinity. So the upper limit becomes \(\infty\).

Substitute these into the integral:

\(\mathop \smallint \limits_0^\pi \frac{1}{{1 + \sin x}}dx = \mathop \smallint \limits_0^\infty \frac{1}{{1 + \frac{2t}{1+t^2}}} \cdot \frac{2dt}{1+t^2}\)

Simplify the denominator:

\(1 + \frac{2t}{1+t^2} = \frac{1+t^2}{1+t^2} + \frac{2t}{1+t^2} = \frac{1+t^2+2t}{1+t^2} = \frac{(1+t)^2}{1+t^2}\)

Substitute the simplified denominator back into the integral:

\(\mathop \smallint \limits_0^\infty \frac{1}{{\frac{(1+t)^2}{1+t^2}}} \cdot \frac{2dt}{1+t^2} = \mathop \smallint \limits_0^\infty \frac{1+t^2}{(1+t)^2} \cdot \frac{2dt}{1+t^2}\)

Cancel out the \((1+t^2)\) terms:

\(= \mathop \smallint \limits_0^\infty \frac{2}{(1+t)^2} dt\)

Evaluate this integral:

\(= 2 \mathop \smallint \limits_0^\infty (1+t)^{-2} dt\)

\(= 2 \left[ \frac{(1+t)^{-1}}{-1} \right]_0^\infty\)

\(= 2 \left[ -\frac{1}{1+t} \right]_0^\infty\)

\(= 2 \left( \lim_{t \to \infty} \left(-\frac{1}{1+t}\right) - \left(-\frac{1}{1+0}\right) \right)\)

\(= 2 \left( 0 - (-1) \right)\)

\(= 2 (1) = 2\)

So, the value of the definite integral is 2.

Final Result for I1 + I2

Now substitute the value of the definite integral back into the expression for \(I_1 + I_2\):

\(I_1 + I_2 = \pi \mathop \smallint \limits_0^\pi \frac{1}{{1 + \sin x}}dx = \pi \times 2\)

\(I_1 + I_2 = 2\pi\)

Thus, the value of \(I_1 + I_2\) is \(2\pi\).

Revision Table: Key Concepts for Integral Evaluation

Additional Information: Solving Trigonometric Integrals

Solving integrals involving trigonometric functions often requires various techniques:

• Using Identities: Simplifying the integrand using identities like \(\sin^2 x + \cos^2 x = 1\), double angle formulas, or sum-to-product formulas can make integration easier.
• Substitution: Simple substitution (like \(u = \sin x\) or \(u = \cos x\)) or more complex ones like the tangent half-angle substitution \(t = \tan(x/2)\) are very useful.
• Integration by Parts: For integrals involving products of functions like \(x \sin x\) or \(x \cos x\).
• Partial Fractions: After a substitution like \(t = \tan(x/2)\), the integrand might become a rational function that can be integrated using partial fraction decomposition.

The integral \(\mathop \smallint \frac{1}{1 + \sin x} dx\) is a classic example where multiplying the numerator and denominator by the conjugate (\(1 - \sin x\)) or using the tangent half-angle substitution works well. Over the interval \(0\) to \(\pi\), the tangent half-angle substitution handles the singularity at \(x=\pi/2\) effectively by transforming the limit to infinity.