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Integration by Parts

Integration by parts uses the formula ∫u dv=uv-∫v du to integrate a product of two functions, with u chosen so that its derivative becomes simpler.

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Student-friendly explanation

This method is useful when the integrand is a product such as x e^x, x sinx, logx, or inverse trigonometric functions. The choice of u matters; the ILATE order often helps choose u: Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential.

How to write this in exams

  1. 1

    Start with the exact idea

    Integration by parts uses the formula ∫u dv=uv-∫v du to integrate a product of two functions, with u chosen so that its derivative becomes simpler.

  2. 2

    Then show how to use it

    Identify the product. Choose u using ILATE and ease of differentiation. Put the remaining factor with dx as dv. Find du and v. Substitute into uv-∫vdu. Evaluate the remaining integral. Differentiate the final answer to check signs.

  3. 3

    Add one concrete example

    For ∫x e^x dx, take u=x and dv=e^x dx. Then du=dx and v=e^x, so the integral is xe^x-∫e^x dx=xe^x-e^x+C.

  4. 4

    Avoid this incomplete answer

    Writing ∫u dv=uv+∫vdu instead of uv-∫vdu, causing a sign error in nearly every by-parts solution.

Definition

Integration by parts uses the formula ∫u dv=uv-∫v du to integrate a product of two functions, with u chosen so that its derivative becomes simpler.

Example

For ∫x e^x dx, take u=x and dv=e^x dx. Then du=dx and v=e^x, so the integral is xe^x-∫e^x dx=xe^x-e^x+C.

Rule to remember

Formula: ∫u dv=uv-∫v du. Conditions: the integrand should be expressible as a product u·dv; v must be integrable; du should simplify the next integral. For ∫logx dx or ∫tan^(-1)x dx, take dv=dx.

Memory hook

First function as u, second function integrated as v, then subtract the new integral.

Examples and method

Worked example

Evaluate ∫x cosx dx. Let u=x, dv=cosx dx. Then du=dx and v=sinx. Using ∫u dv=uv-∫v du, integral =x sinx-∫sinx dx=x sinx+cosx+C. Check: derivative of x sinx+cosx is sinx+xcosx-sinx=xcosx.

Method to apply

Identify the product. Choose u using ILATE and ease of differentiation. Put the remaining factor with dx as dv. Find du and v. Substitute into uv-∫vdu. Evaluate the remaining integral. Differentiate the final answer to check signs.

Diagram support

No diagram is required. A two-column layout for u, dv, du, and v is the most helpful written support.

How CBSE asks it

Common in long-answer questions involving products, logarithmic functions, inverse trigonometric functions, and repeated application such as ∫e^x sinx dx.

Avoid common mistakes

Common confusion

A common mistake is choosing u poorly, making the remaining integral harder instead of simpler.

Common wrong answer

Writing ∫u dv=uv+∫vdu instead of uv-∫vdu, causing a sign error in nearly every by-parts solution.

Exam tip

Use ILATE as a guide, but still check whether differentiating u simplifies the expression and integrating dv is easy.

Quick check

In ∫x sinx dx, what should be chosen as u by ILATE?

Choose u=x, because algebraic functions come before trigonometric functions in ILATE.

Answer writing and exam use

1-mark answer

Integration by parts uses the formula ∫u dv=uv-∫v du to integrate a product of two functions, with u chosen so that its derivative becomes simpler.

2-mark answer

Integration by parts uses the formula ∫u dv=uv-∫v du to integrate a product of two functions, with u chosen so that its derivative becomes simpler. Formula: ∫u dv=uv-∫v du. Conditions: the integrand should be expressible as a product u·dv; v must be integrable; du should simplify the next integral. For ∫logx dx or ∫tan^(-1)x dx, take dv=dx. For ∫x e^x dx, take u=x and dv=e^x dx. Then du=dx and v=e^x, so the integral is xe^x-∫e^x dx=xe^x-e^x+C.

3-mark answer

This method is useful when the integrand is a product such as x e^x, x sinx, logx, or inverse trigonometric functions. The choice of u matters; the ILATE order often helps choose u: Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential. Formula: ∫u dv=uv-∫v du. Conditions: the integrand should be expressible as a product u·dv; v must be integrable; du should simplify the next integral. For ∫logx dx or ∫tan^(-1)x dx, take dv=dx. Evaluate ∫x cosx dx. Let u=x, dv=cosx dx. Then du=dx and v=sinx. Using ∫u dv=uv-∫v du, integral =x sinx-∫sinx dx=x sinx+cosx+C. Check: derivative of x sinx+cosx is sinx+xcosx-sinx=xcosx. Common in long-answer questions involving products, logarithmic functions, inverse trigonometric functions, and repeated application such as ∫e^x sinx dx. Writing ∫u dv=uv+∫vdu instead of uv-∫vdu, causing a sign error in nearly every by-parts solution.
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