Indefinite Integration

Integration by Parts

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Integration by Parts

Integration by Parts: Mastering the Art of Strategic Integration

Hey JEE aspirants! Integration by parts is a powerful technique that unlocks a whole new world of integrals, especially those involving products of functions. Think of it as the integration world's version of the product rule for differentiation. It's essential for tackling many JEE problems, so let's dive in and master it!

The Core Idea: Undoing the Product Rule

Remember the product rule for differentiation? (uv)=uv+uv(uv)' = u'v + uv' Integration by parts is essentially the reverse process of this rule. We aim to strategically break down a complex integral into simpler parts that we can handle more easily. The magic lies in choosing the right parts!

The Integration by Parts Formula

This is the cornerstone of the technique:

udv=uvvdu∫u dv = uv - ∫v du

Let's break it down:

  • uu is a function we choose to differentiate.
  • dvdv is the remaining part of the integrand, which we choose to integrate.
  • dudu is the derivative of uu.
  • vv is the integral of dvdv.

The goal is to choose uu and dvdv such that the integral on the right-hand side, vdu∫v du, is simpler than the original integral, udv∫u dv.

The LIATE Rule: Your Guide to Choosing u

Selecting the right uu is crucial for making integration by parts effective. The LIATE rule provides a helpful hierarchy:

LIATE priority: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential

This means:

  • Logarithmic functions (e.g., ln(x)ln(x), loga(x)log_a(x)) are usually your best choice for uu.
  • Inverse Trigonometric functions (e.g., sin1(x)sin^{-1}(x), tan1(x)tan^{-1}(x)) are next in line.
  • Algebraic functions (e.g., x2x^2, 3x+13x + 1) follow.
  • Trigonometric functions (e.g., sin(x)sin(x), cos(x)cos(x)).
  • Exponential functions (e.g., exe^x, 2x2^x) are typically chosen as dvdv.

The function that comes first in the LIATE order should be your choice for uu. For example, in xsin(x)dx∫x sin(x) dx, xx is algebraic and sin(x)sin(x) is trigonometric. Since algebraic comes before trigonometric in LIATE, we choose u=xu = x and dv=sin(x)dxdv = sin(x) dx.

Let's look at an example to see LIATE in action. Consider xexdx∫x \cdot e^x dx. Here, xx is an algebraic function, and exe^x is an exponential function. According to LIATE, we choose u=xu = x and dv=exdxdv = e^x dx. Then, du=dxdu = dx and v=exdx=exv = ∫e^x dx = e^x. Applying the integration by parts formula:

xexdx=xexexdx=xexex+C∫x e^x dx = x e^x - ∫e^x dx = x e^x - e^x + C

Repeated Integration by Parts

Sometimes, even after applying integration by parts once, the new integral vdu∫v du is still not easily solvable. In such cases, you might need to apply integration by parts again. This is called repeated integration by parts.

Consider x2exdx∫x^2 e^x dx. Let u=x2u = x^2 and dv=exdxdv = e^x dx. Then, du=2xdxdu = 2x dx and v=exv = e^x.

x2exdx=x2ex2xexdx∫x^2 e^x dx = x^2 e^x - ∫2x e^x dx

Now, we have a new integral 2xexdx∫2x e^x dx. We apply integration by parts again, with u=2xu = 2x and dv=exdxdv = e^x dx. Then, du=2dxdu = 2 dx and v=exv = e^x.

2xexdx=2xex2exdx=2xex2ex+C∫2x e^x dx = 2x e^x - ∫2 e^x dx = 2x e^x - 2e^x + C

Substituting this back into our original equation:

x2exdx=x2ex(2xex2ex)+C=x2ex2xex+2ex+C∫x^2 e^x dx = x^2 e^x - (2x e^x - 2e^x) + C = x^2 e^x - 2x e^x + 2e^x + C

The Tabular Method: A Shortcut for Repeated Integration

When dealing with repeated integration by parts, especially when uu is a polynomial, the tabular method can significantly simplify the process. Here's how it works:

  1. Create three columns: Sign, uu (differentiate), and dvdv (integrate).
  2. In the uu column, write your initial uu and differentiate it repeatedly until you reach zero.
  3. In the dvdv column, write your initial dvdv and integrate it repeatedly the same number of times as you differentiated uu.
  4. In the Sign column, alternate signs starting with ++.
  5. Multiply terms diagonally, following the signs.

Let's revisit the example x2exdx∫x^2 e^x dx using the tabular method:

Sign uu (Differentiate) dvdv (Integrate)
+ x2x^2 exe^x
- 2x2x exe^x
+ 22 exe^x
- 00 exe^x

Now, multiply diagonally with the corresponding signs:

x2exdx=+(x2)(ex)(2x)(ex)+(2)(ex)+C=x2ex2xex+2ex+C∫x^2 e^x dx = + (x^2)(e^x) - (2x)(e^x) + (2)(e^x) + C = x^2 e^x - 2x e^x + 2e^x + C

See? Same result, but often faster and less prone to errors!

Tip: The tabular method is especially useful when you have a polynomial multiplied by a trigonometric or exponential function.

Common Mistakes to Avoid

Mistake: Forgetting the constant of integration, CC. Always add it after performing an indefinite integral.
Mistake: Incorrectly applying the LIATE rule. Double-check your choices for uu and dvdv.
Mistake: Making errors in differentiation or integration. Accuracy is key!

JEE-Specific Tricks (When Applicable)

While there isn't a single "magic trick" for integration by parts in JEE, here's a useful approach:

  1. Look for patterns: JEE often tests you on integrals that require clever manipulation before applying integration by parts.
  2. Consider substitutions: Sometimes, a substitution can simplify the integral, making integration by parts easier or even unnecessary.
  3. Practice, practice, practice: The more problems you solve, the better you'll become at recognizing when and how to apply integration by parts effectively.

Keep practicing, and you'll become a master of integration by parts! All the best for your JEE preparation.