Calculate Integral I2 Value

The problem asks us to find the value of the definite integral \(I_2\), which is given by:

\(I_2=\int_0^\pi \frac{1}{1+\sin ^2 x} d x\)

Step-by-Step Evaluation of Integral I2

To evaluate \(I_2\), we can use properties of definite integrals.

The integrand is \(f(x) = \frac{1}{1+\sin^2 x}\). Let's check the property \(f(\pi - x)\).

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

Since \(\sin(\pi - x) = \sin x\), we have:

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

Since \(f(\pi - x) = f(x)\), we can use the property \(\int_0^{2a} f(x) dx = 2 \int_0^a f(x) dx\), where \(2a = \pi\), so \(a = \pi/2\).

Therefore, \(I_2\) can be written as:

\(I_2 = 2 \int_0^{\pi/2} \frac{1}{1+\sin ^2 x} d x\)

Now, let's evaluate the integral \(J = \int_0^{\pi/2} \frac{1}{1+\sin ^2 x} d x\).

We can divide the numerator and denominator by \(\cos^2 x\):

\(J = \int_0^{\pi/2} \frac{\frac{1}{\cos^2 x}}{\frac{1+\sin^2 x}{\cos^2 x}} d x = \int_0^{\pi/2} \frac{\sec^2 x}{\frac{1}{\cos^2 x} + \frac{\sin^2 x}{\cos^2 x}} d x\)

\(J = \int_0^{\pi/2} \frac{\sec^2 x}{\sec^2 x + \tan^2 x} d x\)

Substitute \(\sec^2 x = 1 + \tan^2 x\) in the denominator:

\(J = \int_0^{\pi/2} \frac{\sec^2 x}{(1+\tan^2 x) + \tan^2 x} d x = \int_0^{\pi/2} \frac{\sec^2 x}{1 + 2\tan^2 x} d x\)

Now, let's use the substitution method. Let \(t = \tan x\). Then, \(dt = \sec^2 x \, dx\).

We need to change the limits of integration based on the substitution:

• When \(x = 0\), \(t = \tan(0) = 0\).
• When \(x = \pi/2\), \(t = \tan(\pi/2)\), which approaches \(\infty\).

The integral \(J\) becomes:

\(J = \int_0^\infty \frac{1}{1 + 2t^2} dt\)

To evaluate this integral, we can factor out 2 from the denominator:

\(J = \int_0^\infty \frac{1}{2(\frac{1}{2} + t^2)} dt = \frac{1}{2} \int_0^\infty \frac{1}{(\frac{1}{\sqrt{2}})^2 + t^2} dt\)

This is in the form \(\int \frac{1}{a^2 + x^2} dx = \frac{1}{a} \arctan(\frac{x}{a})\). Here \(a = \frac{1}{\sqrt{2}}\) and the variable is \(t\).

\(J = \frac{1}{2} \left[ \frac{1}{1/\sqrt{2}} \arctan\left(\frac{t}{1/\sqrt{2}}\right) \right]_0^\infty\)

\(J = \frac{1}{2} \left[ \sqrt{2} \arctan\left(\sqrt{2}t\right) \right]_0^\infty\)

\(J = \frac{\sqrt{2}}{2} \left[ \arctan\left(\sqrt{2}t\right) \right]_0^\infty\)

\(J = \frac{1}{\sqrt{2}} \left( \lim_{t \to \infty} \arctan\left(\sqrt{2}t\right) - \arctan\left(\sqrt{2} \times 0\right) \right)\)

\(J = \frac{1}{\sqrt{2}} \left( \arctan(\infty) - \arctan(0) \right)\)

We know that \(\arctan(\infty) = \pi/2\) and \(\arctan(0) = 0\).

\(J = \frac{1}{\sqrt{2}} \left( \frac{\pi}{2} - 0 \right) = \frac{1}{\sqrt{2}} \frac{\pi}{2} = \frac{\pi}{2\sqrt{2}}\)

Now we need to find \(I_2\), which is \(2J\).

\(I_2 = 2 \times J = 2 \times \frac{\pi}{2\sqrt{2}} = \frac{\pi}{\sqrt{2}}\)

Thus, the value of \(I_2\) is \(\frac{\pi}{\sqrt{2}}\).

Comparing with Options

Let's compare our calculated value of \(I_2\) with the given options:

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

Our calculated value \(\frac{\pi}{\sqrt{2}}\) matches Option 1.

Revision Table: Key Concepts Used for Integral I2

Additional Information: Understanding Trigonometric Integrals

Integrals involving trigonometric functions often require specific techniques. For integrals with \(\sin^n x \cos^m x\), different strategies apply based on the parity of \(n\) and \(m\). However, for integrands like \(\frac{1}{a + b \sin^2 x}\) or \(\frac{1}{a + b \cos^2 x}\) or \(\frac{1}{a \sin^2 x + b \cos^2 x + c}\), a common approach is to divide the numerator and denominator by \(\cos^2 x\).

Dividing by \(\cos^2 x\) transforms the integrand into a rational function of \(\tan x\) and \(\sec^2 x\). Since \(\sec^2 x\) is the derivative of \(\tan x\), this transformation prepares the integral for a simple substitution \(t = \tan x\). Remember that \(\sec^2 x = 1 + \tan^2 x\) should be used to express any remaining \(\sec^2 x\) terms in the denominator in terms of \(\tan x\).

After the substitution \(t = \tan x\), the integral becomes an integral of a rational function of \(t\). These integrals can often be solved using standard integration formulas, partial fraction decomposition, or other algebraic techniques. The definite integral limits must always be changed according to the substitution used.

In the case of \(I_2 = \int_0^\pi \frac{1}{1+\sin ^2 x} d x\), using the property \(\int_0^{2a} f(x) dx = 2 \int_0^a f(x) dx\) when applicable is crucial as it changes the upper limit from \(\pi\) to \(\pi/2\), which corresponds to \(\tan x\) covering the range from 0 to \(\infty\) exactly once, making the substitution \(t=\tan x\) straightforward over the interval \([0, \pi/2]\).