Solving Integrals Through Substitution and Simplification in Calculus

Calculus, a fundamental branch of mathematics, involves solving a variety of problems through integration and differentiation. One powerful technique in integration is making appropriate substitutions to simplify the integrand. In this article, we will explore how to integrate a given function by utilizing substitution and simplify the final result. This process is not only beneficial in mathematics but also finds applications in physics, engineering, and other scientific fields.

Introduction to the Integral Problem

We will consider the integral of the form:

Integral: [int_0^{frac{pi}{2}} cos(t) sin^3(t) sqrt{14cos^2(t)} , dt]

Step-by-Step Solution

Step 1: Simplify the Radicand

First, we simplify the radicand (the expression inside the square root) by expressing (cos^2(t)) in terms of (sin^2(t)). Recall that (cos^2(t) 1 - sin^2(t)). Substituting this into our integral, we get:

Simplified Form: [int_0^{frac{pi}{2}} cos(t) sin^3(t) sqrt{5 - 4sin^2(t)} , dt]

Step 2: Use Substitution

For further simplification, let's make the substitution (u sin^2(t)). Then, (du 2sin(t)cos(t) , dt). By manipulating the integral, we can integrate with respect to (u).

Substituted Integral: [frac{1}{2} int_0^1 u sqrt{5 - 4u} , du]

Step 3: Further Simplification

To integrate the expression (u sqrt{5 - 4u}), we attempt to convert (u) outside the square root. Notice that (u frac{1}{4}(5 - (5 - 4u))). Thus, we can write:

Integral Transformation: [frac{1}{2} int_0^1 u sqrt{5 - 4u} , du frac{1}{8} left[ int_0^1 5sqrt{5 - 4u} , du - int_0^1 (5 - 4u) sqrt{5 - 4u} , du right]]

Evaluation of Integrals: [frac{1}{8} left[ -frac{5}{6} left(5 - 4uright)^{frac{3}{2}} bigg|_0^1 - frac{1}{10} left(5 - 4uright)^{frac{5}{2}} bigg|_0^1 right]]

Final Result: [frac{1}{8} left[ -frac{5}{6} left(5^frac{3}{2} - 1right) frac{1}{10} left(5^frac{5}{2} - 1right) right] boxed{frac{1}{120} left(5^frac{5}{2} - 11right)}]

Integration by Substitution: A General Method

Substitution is a powerful technique for solving integrals, and it often simplifies complex expressions. In the above steps, we first used the identity (cos^2(t) 1 - sin^2(t)) to simplify the radicand. Then, by setting (u sin^2(t)), we transformed the integral into a more manageable form. This method is particularly useful when dealing with functions involving square roots, trigonometric identities, and other complex expressions.

Conclusion

Using substitution to solve integrals can greatly simplify and streamline the process. By carefully choosing the substitution, we can transform complex integrals into simpler forms, making them easier to evaluate. This technique is not only useful in calculus but also has broad applications in various fields of science and engineering.

Keywords: integral, substitution, calculus