Problem 5 - Calculus 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

OR \(\int_{0}^{\pi / 2} \sin ^{n} x \cos ^{n} x d x\)

Step-by-Step Solution

Answer

The integral involves symmetry with specific transformation simplifications, leading to evaluations often with known symmetric properties or special functions.

Step 1: Recognize the Symmetry

Notice that both functions, \(\sin(x)\) and \(\cos(x)\), are symmetric and periodic over the interval \([0, \pi/2]\). This symmetry can sometimes allow us to simplify integrals involving powers of these functions.

Step 2: Use Trigonometric Identity

Apply the identity \( \sin(2x) = 2\sin(x)\cos(x) \) to express the product \( \sin^n(x)\cos^n(x) \) in terms of \(\sin(2x)\). So, we have \( \sin^n(x)\cos^n(x) = \left(\frac{1}{2}\sin(2x)\right)^n \).

Step 3: Simplify the Integral

The integral becomes \(\int_{0}^{\pi/2} \left(\frac{1}{2}\right)^n (\sin(2x))^n dx\). Notice that the integral limits change when simplifying, but they remain the same in this transformation.

Step 4: Substitute and Simplify Again

Set \(u = 2x\), hence \(du = 2dx\). The limits of integration change accordingly: when \(x = 0\), \(u = 0\) and when \(x = \pi/2\), \(u = \pi\). The integral now becomes \(\frac{1}{2}\int_{0}^{\pi} \left(\frac{1}{2}\sin(u)\right)^n \cdot \frac{1}{2} du \).

Step 5: Evaluate the Integral

Now evaluate the simplified integral \(\frac{1}{2^{n+1}} \int_{0}^{\pi} \sin^n(u) du\). This is often a known integral that can be simplified through reduction formulas or symmetry properties for specific cases of \(n\). For the generic case, you'd follow specific recurrence formulas for even/odd \(n\).

Step 6: Apply Detailed Formula/Recurrence

For some specific values of \(n\), use the beta function, which relates to symmetric power functions: \[ \int_0^\pi \sin^n u \, du = \frac{\sqrt{\pi} \Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n}{2} + 1\right)} \]. Use this for deeper calculations if \(n\) were specified, adjusting by symmetry and periodic properties as necessary.

Key Concepts

Trigonometric Identities

Trigonometric identities are essential tools in simplifying integrals, especially when dealing with powers of trigonometric functions. They allow us to rewrite complex expressions into more manageable forms. In this exercise, the identity \( \sin(2x) = 2\sin(x)\cos(x) \) was used. By recognizing that \( \sin^n(x)\cos^n(x) \) can be transformed into \( \left(\frac{1}{2}\sin(2x)\right)^n \), we leverage this formula to simplify the original integral.

• These identities help reduce the complexity of the integral.
• They provide a pathway to use simpler substitution or direct integration techniques.
• Understanding such identities is crucial as they appear frequently in calculus problems.

An important takeaway is recognizing when and how these identities can be applied to modify the structure of the problem, leading to more straightforward solutions.

Symmetry in Integration

Symmetry plays a key role when evaluating integrals over specific intervals. The integral \( \int_{0}^{\pi / 2} \sin^n(x)\cos^n(x) dx \) lies over a symmetric interval where both \( \sin(x) \) and \( \cos(x) \) are symmetric functions. This property can significantly simplify calculations.

• In symmetric intervals, the behavior of trigonometric functions can reduce complex calculations.
• Symmetry can also assist in establishing bounds which remain consistent even after substitutions.

In this case, understanding trigonometric symmetry along the interval \([0, \pi/2]\) is key as it suggest intuitive strategies for solving these integrals without over-complicating the process.

Gamma Function

The Gamma Function \( \Gamma(n) \) is a mathematical concept that extends the factorial function to complex numbers. It is a powerful tool often used in calculus, particularly for solving integrals involving trigonometric functions raised to a power. In the integral \( \int_{0}^{\pi} \sin^n(u) du \), the Gamma function helps facilitate solutions for integrals by reducing them into "standard forms".

• The Gamma Function relates to the factorial function by \( \Gamma(n+1) = n! \) for natural numbers \( n \).
• Makes handling integrals with various power rules more manageable.
• Offers a bridge between trigonometric identities and factorial math.

The insight is to leverage it for reducing complex integral calculations into known results, especially for any repeating or recursive formulas.

Substitution Method

The substitution method in calculus helps us transform an integral into a simpler form, often making it straightforward to evaluate. In this exercise, the substitution \( u = 2x \) was employed, which helps manage the integral limits and the function itself by simplifying \( \int (\sin(2x))^n dx \) into \( \int \sin^n(u) du \).

• Substitution is an essential method which allows flexible adjustment of variables.
• Correctly adjusting limits of integration is vital after a substitution.
• Consideration of symmetry and compatibility of substitution can optimize results.

This method is important for students to master, as it turns complex integral problems into ones that can be handled with more direct integration techniques, demonstrating the power of variable changes strategically aligned with problem symmetry and identity.