Solving Integrals ∫(x+1)^2 E^(3x) Dx And ∫cos(ln X) Dx A Comprehensive Guide
In the realm of calculus, integration stands as a fundamental operation, serving as the inverse of differentiation. Mastering integration techniques is crucial for solving a wide array of problems in mathematics, physics, engineering, and other scientific disciplines. This article delves into the intricacies of evaluating two specific integrals: ∫(x+1)^2 e^(3x) dx and ∫cos(ln x) dx. These integrals exemplify the application of different integration methods, including integration by parts and the use of substitutions. We will explore each integral step-by-step, providing a comprehensive understanding of the solution process and the underlying principles.
To tackle the integral ∫(x+1)^2 e^(3x) dx, we employ the powerful technique of integration by parts. This method is particularly effective when dealing with integrals involving the product of two functions. The integration by parts formula is given by:
∫u dv = uv - ∫v du
where u and v are functions of x, and du and dv are their respective differentials. The key to successfully applying integration by parts lies in choosing appropriate functions for u and dv. A helpful guideline is the acronym LIATE, which prioritizes functions in the order of Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential. This order suggests which function to choose as u, with the higher priority functions being preferred.
In our case, we have the product of an algebraic function (x+1)^2 and an exponential function e^(3x). Following the LIATE rule, we choose (x+1)^2 as u and e^(3x) dx as dv. This gives us:
u = (x+1)^2 dv = e^(3x) dx
Now, we need to find du and v. Differentiating u with respect to x, we get:
du = 2(x+1) dx
Integrating dv with respect to x, we obtain:
v = ∫e^(3x) dx = (1/3)e^(3x)
Applying the integration by parts formula, we have:
∫(x+1)^2 e^(3x) dx = (x+1)^2 (1/3)e^(3x) - ∫(1/3)e^(3x) 2(x+1) dx
Simplifying the expression, we get:
∫(x+1)^2 e^(3x) dx = (1/3)(x+1)^2 e^(3x) - (2/3)∫(x+1)e^(3x) dx
We now encounter another integral, ∫(x+1)e^(3x) dx, which again requires integration by parts. We repeat the process, this time choosing:
u = x+1 dv = e^(3x) dx
Differentiating u, we get:
du = dx
Integrating dv, we obtain:
v = (1/3)e^(3x)
Applying integration by parts again:
∫(x+1)e^(3x) dx = (x+1)(1/3)e^(3x) - ∫(1/3)e^(3x) dx
Simplifying:
∫(x+1)e^(3x) dx = (1/3)(x+1)e^(3x) - (1/3)∫e^(3x) dx
Now, the integral ∫e^(3x) dx is straightforward:
∫e^(3x) dx = (1/3)e^(3x) + C
Substituting this back into the previous equation:
∫(x+1)e^(3x) dx = (1/3)(x+1)e^(3x) - (1/9)e^(3x) + C
Finally, substituting this result back into our original equation:
∫(x+1)^2 e^(3x) dx = (1/3)(x+1)^2 e^(3x) - (2/3)[(1/3)(x+1)e^(3x) - (1/9)e^(3x)] + C
Expanding and simplifying, we arrive at the final solution:
∫(x+1)^2 e^(3x) dx = (1/3)(x+1)^2 e^(3x) - (2/9)(x+1)e^(3x) + (2/27)e^(3x) + C
This meticulous step-by-step process demonstrates the application of integration by parts, highlighting the importance of choosing the appropriate functions for u and dv. The repeated application of the technique showcases its versatility in handling complex integrals.
For the integral ∫cos(ln x) dx, a different approach is required. We start by employing a u-substitution. Let's substitute:
u = ln x
Then, differentiating both sides with respect to x, we get:
du = (1/x) dx
Rearranging, we have:
dx = x du
Since u = ln x, we can express x in terms of u as:
x = e^u
Substituting these into the original integral:
∫cos(ln x) dx = ∫cos(u) e^u du
Now, we have a new integral, ∫e^u cos(u) du, which can be solved using integration by parts. This integral is a classic example where integration by parts needs to be applied twice, and a clever algebraic manipulation is used to solve for the integral.
Let's apply integration by parts. First, we choose:
p = e^u dq = cos(u) du
Then,
dp = e^u du q = ∫cos(u) du = sin(u)
Applying the integration by parts formula:
∫e^u cos(u) du = e^u sin(u) - ∫sin(u) e^u du
Now, we have another integral, ∫sin(u) e^u du, which again requires integration by parts. We choose:
r = e^u ds = sin(u) du
Then,
dr = e^u du s = ∫sin(u) du = -cos(u)
Applying integration by parts again:
∫sin(u) e^u du = -e^u cos(u) - ∫(-cos(u)) e^u du
∫sin(u) e^u du = -e^u cos(u) + ∫cos(u) e^u du
Substituting this back into our previous equation:
∫e^u cos(u) du = e^u sin(u) - [-e^u cos(u) + ∫cos(u) e^u du]
∫e^u cos(u) du = e^u sin(u) + e^u cos(u) - ∫e^u cos(u) du
Now, we have the same integral on both sides of the equation. We can move the integral on the right side to the left side:
2∫e^u cos(u) du = e^u sin(u) + e^u cos(u)
Dividing both sides by 2:
∫e^u cos(u) du = (1/2)[e^u sin(u) + e^u cos(u)] + C
Finally, we substitute back u = ln x:
∫cos(ln x) dx = (1/2)[e^(ln x) sin(ln x) + e^(ln x) cos(ln x)] + C
∫cos(ln x) dx = (1/2)[x sin(ln x) + x cos(ln x)] + C
∫cos(ln x) dx = (x/2)[sin(ln x) + cos(ln x)] + C
This solution demonstrates the power of combining substitution and integration by parts. The clever algebraic manipulation to solve for the integral is a key technique in handling integrals of this type. The final result showcases the elegance of the solution process.
In conclusion, the evaluation of ∫(x+1)^2 e^(3x) dx and ∫cos(ln x) dx highlights the versatility and power of integration techniques. The first integral, ∫(x+1)^2 e^(3x) dx, was solved using repeated integration by parts, demonstrating the importance of strategic function selection. The second integral, ∫cos(ln x) dx, required a combination of u-substitution and integration by parts, showcasing the need for adaptability in problem-solving. Mastering these techniques and understanding their applications is essential for anyone delving into the world of calculus and its applications. The ability to choose the appropriate method and execute it effectively is a hallmark of mathematical proficiency. These examples serve as valuable exercises for honing integration skills and building a solid foundation in calculus.
