AP Calculus AB/BC28 views·Updated Jun 11, 2026·2 pages

Integration by Parts in Calculus BC

Integration by parts is a powerful technique for solving complex...

of 2

$9/15173 Integration by Parts $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$ $d(uv) = udv + vdu$ $d(uv) - vdu = udv$ $udv = d(uv$

Integration by Parts Formula

Integration by parts uses the formula ∫u dv = uv - ∫v du, which comes from rearranging the product rule for derivatives. This technique is especially useful when you have a product of functions and direct integration seems impossible.

When choosing which function to call "u," remember the helpful acronym "LIATE" which stands for Logarithmic, Inverse trigonometric, Algebraic (polynomials), Trigonometric, and Exponential functions. Functions earlier in this list make better choices for "u."

For example, to solve ∫ $3x+1$ (sin 5x)dx, we set u = 3x+1 and dv = sin 5x dx. After finding du = 3dx and v = -⅕cos 5x, we substitute into our formula: uv - ∫v du. This gives us -⅕ $3x+1$ (cos 5x) + ⅗∫cos 5x dx, which simplifies to -⅕ $3x+1$ (cos 5x) + ⅗(⅕sin 5x) + c.

💡 Pro Tip: When choosing which function to be "u" and which to be "dv," pick the function for "u" that simplifies when differentiated, and the function for "dv" that remains manageable when integrated.

of 2

$9/15173 Integration by Parts $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$ $d(uv) = udv + vdu$ $d(uv) - vdu = udv$ $udv = d(uv$

Tabular Integration Method

The tabular method is a shortcut for integration by parts when you need to apply the technique multiple times. It's especially useful for integrals involving products of polynomials with trigonometric or exponential functions.

To use this method, arrange the functions in a table with derivatives of one function down the left side and integrals of the other function down the right. Continue until the left column reaches zero. Then multiply diagonally, alternating plus and minus signs, to get your answer.

For example, with ∫x³sin x dx, we put x³ on the left (since it differentiates to zero eventually) and sin x on the right. We differentiate x³ repeatedly until we reach 0, and integrate sin x repeatedly. The final answer is -x³cos x + 3x²sin x - 6x cos x - 6sin x + C.

🔑 Remember: The tabular method saves time when integrating products where one function differentiates to zero after several steps (like polynomials) and the other has simple repeated integrals $like sin/cos or exponentials$ .

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AP Calculus AB/BC28 views·Updated Jun 11, 2026·2 pages

Integration by Parts in Calculus BC

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Integration by parts is a powerful technique for solving complex integrals when simple substitution won't work. It's based on the product rule for derivatives but used in reverse, allowing you to transform difficult integrals into simpler ones that you can...

of 2

$9/15173 Integration by Parts $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$ $d(uv) = udv + vdu$ $d(uv) - vdu = udv$ $udv = d(uv$

Integration by Parts Formula

💡 Pro Tip: When choosing which function to be "u" and which to be "dv," pick the function for "u" that simplifies when differentiated, and the function for "dv" that remains manageable when integrated.

of 2

$9/15173 Integration by Parts $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$ $d(uv) = udv + vdu$ $d(uv) - vdu = udv$ $udv = d(uv$

Tabular Integration Method

🔑 Remember: The tabular method saves time when integrating products where one function differentiates to zero after several steps (like polynomials) and the other has simple repeated integrals $like sin/cos or exponentials$ .