Introduction:

Integration by Parts is a crucial technique in integral calculus, especially useful when dealing with the integration of the product of two functions. It is derived from the product rule of differentiation and helps solve complex integrals that cannot be integrated directly.

Understanding this method thoroughly will aid in solving various problems in CBSE Class 12 board exams and competitive exams like JEE Main.

Formula for Integration by Parts:

If u = f(x) and v = g(x), then:

∫ u·v dx = u ∫v dx - ∫ (du/dx · ∫v dx) dx

Or simply,

∫ u·v dx = uv - ∫ v·(du/dx) dx

Choosing u and v – ILATE Rule:

To select which function to differentiate and which to integrate, use the ILATE rule:

I: Inverse Trigonometric functions

L: Logarithmic functions

A: Algebraic functions

T: Trigonometric functions

E: Exponential functions

Solved Examples of Integration by Parts:

Evaluate ∫ x · e^x dx

Let u = x (Algebraic), dv = e^x dx

Then, du = dx, and v = ∫ e^x dx = e^x

Apply the formula:

∫ x·e^x dx = x·e^x - ∫ e^x dx = x·e^x - e^x + C

Answer: ∫ x·e^x dx = e^x(x - 1) + C

Evaluate ∫ ln x dx

Let u = ln x, dv = dx

Then, du = (1/x) dx, v = ∫ dx = x

Apply the formula:

∫ ln x dx = x·ln x - ∫ x·(1/x) dx = x·ln x - ∫ 1 dx = x·ln x - x + C

Answer: ∫ ln x dx = x(ln x - 1) + C

Evaluate ∫ x · sin x dx

Let u = x, dv = sin x dx

Then, du = dx, v = ∫ sin x dx = -cos x

Apply the formula:

∫ x·sin x dx = -x·cos x + ∫ cos x dx = -x·cos x + sin x + C

Answer: ∫ x·sin x dx = -x·cos x + sin x + C

Special Cases and Tips:

Some integrals may require repeated application of the formula. For example:

∫ x^2 e^x dx

Practice Questions

1. ∫ x · cos x dx

2. ∫ x · ln x dx

3. ∫ x^2 · e^x dx

4. ∫ arctan x dx

5. ∫ ln x dx

Conclusion:

Integration by Parts is a powerful technique in calculus, especially when dealing with products of functions. Mastery of the ILATE rule and regular practice of varied problems ensures confidence and accuracy in the exams.