Evaluating and Integrating 1/32sinx from 0 to π

In the realm of calculus, evaluating integrals can often be a challenging task, especially when dealing with trigonometric functions. One such example is the integral of 1/32sinx from 0 to π. Let's explore the process and methods used to solve such integrals systematically.

Introduction to the Problem

Consider the integral:

#8747;0π (1/32sinx) dx.

The problem at hand is to evaluate this integral using appropriate substitutions and techniques. One helpful approach is provided by Weierstrass substitution, which transforms trigonometric functions into algebraic ones, making the integral more manageable.

Applying Weierstrass Substitution

Let's implement the Weierstrass substitution, where t tan(x/2). This substitution is particularly useful for transforming integrals involving trigonometric functions into rational functions.

Step-by-Step Solution

Substitute t tan(x/2) into the integral: [dx frac{2}{1 t^2} dt] [sin(x) frac{2t}{1 t^2}] Thus, the integral becomes: [∫_{0}^{∞} frac{1}{32 cdot frac{2t}{1 t^2}} cdot frac{2}{1 t^2} dt] Simplify the expression inside the integral: [ frac{1}{32} cdot frac{2(1 t^2)}{2t(1 t^2)} frac{1}{16t}]

Further Simplification

To make the integral even easier to handle, let's multiply the numerator and denominator by 3:

[ frac{3}{48t} frac{3}{3t^2 12t 9} frac{3}{3(t^2 4t 3)}]

This results in:

[ frac{1}{16} cdot frac{3}{(t 3)(t 1)}]

Fraction Decomposition

Decompose the fraction using partial fractions:

[frac{1}{(t 3)(t 1)} frac{A}{t 3} frac{B}{t 1}]

Solve for A and B:

[frac{1}{(t 3)(t 1)} frac{1/2}{t 3} - frac{1/2}{t 1}]

Substitute back into the integral:

[frac{1}{16} cdot frac{3}{(t 3)(t 1)} frac{3}{32} left( frac{1/2}{t 3} - frac{1/2}{t 1} right)]

Final Integration

Integrate each term separately:

[frac{3}{32} int_{0}^{∞} left( frac{1/2}{t 3} - frac{1/2}{t 1} right) dt]

This becomes:

[frac{3}{32} left[ frac{1}{2} ln|t 3| - frac{1}{2} ln|t 1| right]_{0}^{∞}]

Evaluate the limits:

[frac{3}{32} left[ frac{1}{2} left( ln(t 3) - ln(t 1) right) right]_{0}^{∞}]

[ frac{3}{64} left[ ln left( frac{t 3}{t 1} right) right]_{0}^{∞}]

Limits evaluation:

[ frac{3}{64} left[ ln left( frac{∞ 3}{∞ 1} right) - ln left( frac{0 3}{0 1} right) right]]

[ frac{3}{64} left[ ln(1) - ln(3) right]]

[ -frac{3ln(3)}{64}]

Alternative Approach: Another Integral Form

Consider the integral:

[I ∫_{0}^{π} frac{1}{32 cdot 2csc x} dx]

This can be rewritten using the identity:

[2csc x frac{2}{sin x}]

[I ∫_{0}^{π} frac{sin x}{3sin x - 2} dx]

Using the substitution:

[sin x 2 frac{tan(x/2)}{1 tan^2(x/2)}]

[dx frac{2}{1 tan^2(x/2)} dx]

This leads to:

[I ∫_{0}^{∞} frac{1}{3 cdot frac{2t}{1 t^2} - 2} cdot frac{2}{1 t^2} dt]

[ ∫_{0}^{∞} frac{2}{3 cdot 2t - 2(1 t^2)} dt]

[ ∫_{0}^{∞} frac{2}{6t - 2 - 2t^2} dt]

[ ∫_{0}^{∞} frac{2}{-2(2-3t)(t 1)} dt]

This further simplification allows us to solve the integral using standard techniques.

Conclusion

The integral of 1/32sinx from 0 to π can be solved using Weierstrass substitution and additional algebraic manipulation. This article illustrates the detailed steps and transformations required to evaluate such integrals. Understanding these techniques is crucial for advanced calculus and real-world applications involving trigonometric functions.