Integration By Parts

The Organic Chemistry Tutor
25 Mar 202132:51

Summary

TLDRThis educational video script covers the technique of integration by parts to solve various integral problems. It explains the process step by step, starting with the integral of x times e^x and progressing to more complex examples such as the integral of x^2 ln(x) and combinations of exponential and trigonometric functions. The script also tackles integrals involving logarithmic functions and provides a method to solve definite integrals, exemplified by e^(3x)cos(4x). Each example builds on the previous, reinforcing the concept of integration by parts for students to master calculus integration.

Takeaways

  • 📚 Integration by parts is a fundamental technique for solving integrals of products of functions, expressed as ∫udv = uv - ∫vdu.
  • 🔍 Choosing 'u' and 'dv' wisely is crucial; typically, 'u' is chosen to simplify when differentiated, and 'dv' is chosen to be easily integrable.
  • 📈 The integral of x * e^x is found by setting x as 'u' and e^x dx as 'dv', resulting in x * e^x - e^x + C.
  • 🌀 For integrals involving trigonometric functions, such as x * sin(x), setting x as 'u' and sin(x) dx as 'dv' leads to -x * cos(x) + sin(x) + C.
  • 📘 When integrating a product of a polynomial and a logarithmic function, like x^2 * ln(x), it's often necessary to apply integration by parts multiple times.
  • ⚡ The integral of ln(x) is x * ln(x) - x + C, which is derived using integration by parts and recognizing the integral of 1 dx as x.
  • 🔗 The integral of x^2 * sin(x) involves using integration by parts twice and results in a complex expression involving both sine and cosine functions.
  • 🌟 The integral of x^2 * e^x requires applying integration by parts twice, leading to a solution involving powers of e^x.
  • 📉 The integral of ln(x) squared involves rewriting the expression and applying integration by parts, resulting in x * ln(x) squared - 2x * ln(x) + 2x + C.
  • 📊 The integral of ln(x) to the seventh power is simplified by recognizing it as 7 * ln(x), leading to 7 * (x * ln(x) - x) + C.
  • 🌌 The integral of e^x * sin(x) is solved by recognizing the pattern in the integral and using a clever manipulation to find the solution as (1/2) * e^x * (sin(x) - cos(x)) + C.

Q & A

  • What is the integral of x times e^x?

    -The integral of x times e^x is found using integration by parts, setting u=x and dv=e^x dx, which gives du=dx and v=e^x. The result is x * e^x - integral of e^x dx, which simplifies to x * e^x - e^x + C, where C is the constant of integration.

  • How do you integrate x times sine x dx?

    -For x times sine x, set u=x and dv=sine x dx, giving du=dx and v=-cosine x. Using integration by parts, the integral becomes -x * cosine x + integral of cosine x dx, which simplifies to -x * cosine x + sine x + C.

  • What is the antiderivative of x squared times ln x?

    -To find the antiderivative of x squared times ln x, set dv=x squared dx and u=ln x, giving v=x cubed / 3 and du=1/x dx. The integral by parts results in (1/3) * x cubed * ln x - (1/3) * integral of x squared dx, which simplifies to (1/3) * x cubed * ln x - (1/9) * x cubed + C.

  • How can you find the integral of the natural log of x using integration by parts?

    -For the natural log of x, set u=ln x and dv=dx, giving du=1/x dx and v=x. The integral by parts formula yields x * ln x - integral of x * (1/x) dx, which simplifies to x * ln x - x + C.

  • What is the indefinite integral of x squared times sine x dx?

    -For x squared times sine x, you need to use integration by parts twice. Initially, set u=x squared and dv=sine x dx, giving du=2x dx and v=-cosine x. The process involves further integration by parts on the resulting integral of 2x * cosine x dx, leading to a final answer involving x squared * cosine x, 2x * sine x, and 2 * cosine x, plus the constant C.

  • How do you integrate x squared times e^x dx?

    -For x squared times e^x, use integration by parts twice. Start by setting u=x squared and dv=e^x dx, giving du=2x dx and v=e^x. The process involves another round of integration by parts with u=2x and dv=e^x dx, leading to a final expression involving e^x terms and a constant C.

  • What is the integral of ln x squared dx?

    -To integrate ln x squared, set u=ln x squared and dv=dx, giving du=2 * ln x * (1/x) dx and v=x. The integral by parts results in x * ln x squared - 2 integral of ln x * (1/x) dx. Solving this requires recognizing the integral of ln x and simplifying to get the final answer involving x * ln x squared, -2x * ln x, and 2x, plus C.

  • How can you rewrite the integral of ln x to the seventh power dx?

    -The integral of ln x to the seventh power can be rewritten using the property of logs to move the exponent to the front, resulting in 7 * integral of ln x dx. Since the integral of ln x is x * ln x - x, the final answer is 7 * (x * ln x - x) + C.

  • What is the indefinite integral of e^x times sine x dx?

    -For e^x times sine x, set u=sine x and dv=e^x dx, giving du=cosine x dx and v=e^x. Using integration by parts, the process involves recognizing the similarity of the resulting terms to the original integral and solving for the integral, leading to a final answer involving e^x * (sine x - cosine x) / 2 plus C.

  • How do you find the antiderivative of ln x squared divided by x dx?

    -To find the antiderivative of ln x squared / x, rewrite it as ln x squared * 1/x dx and use integration by parts with u=ln x squared and dv=1/x dx, giving du=2 * ln x * (1/x) dx and v=ln x. The process involves recognizing the similarity of terms and solving for the integral, resulting in ln x cubed / 3 plus C.

  • What is the indefinite integral of e^(3x) times cosine 4x dx?

    -For e^(3x) times cosine 4x, set u=cosine 4x and dv=e^(3x) dx, giving du=-sine 4x * 4 dx and v=(1/3)e^(3x). Using integration by parts, the process involves moving constants to the front and recognizing the similarity of terms, leading to a final answer involving e^(3x) * (cosine 4x / 3 + 4/3 * sine 4x) plus C.

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