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Integrals - Integration of a variety of functions by substitution, by partial fractions and by parts

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Integration by Substitution: This technique involves transforming the variable xx into a new variable tt using a substitution t=g(x)t = g(x), such that the derivative g′(x)g'(x) is also present in the integrand. Visually, this can be interpreted as a coordinate transformation that simplifies a complex curve into a standard form, effectively stretching or compressing the xx-axis to make the area under the curve easier to calculate.

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Method of Partial Fractions: Used for integrating rational functions where the degree of the numerator is less than the degree of the denominator. The fraction is decomposed into a sum of simpler fractions whose integrals are known (usually logarithmic or inverse trigonometric). Geometrically, this is equivalent to breaking down a complex curve with multiple vertical asymptotes into simpler component curves whose individual behaviors are well-understood.

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Integration by Parts: Based on the product rule of differentiation, this method is used when the integrand is a product of two functions, uu and vv. The formula is derived from the area of a rectangle in the u−vu-v plane, where the total area uvuv is the sum of the areas integrated along both axes: ∫u dv+∫v du\int u \, dv + \int v \, du. This allows us to exchange the integral of one product for another, simpler one.

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The ILATE Rule: A priority-based mnemonic used to choose the function uu in integration by parts. The acronym stands for Inverse Trigonometric, Logarithmic, Algebraic, Trigonometric, and Exponential functions. Choosing uu based on this hierarchy ensures that the derivative dudu becomes simpler (or doesn't get more complex), making the second integral ∫v du\int v \, du manageable.

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Special Integrals and Geometric Substitutions: For integrands involving square roots of quadratic expressions like a2−x2\sqrt{a^2 - x^2} or x2+a2\sqrt{x^2 + a^2}, trigonometric substitutions such as x=asin⁡θx = a \sin \theta or x=atan⁡θx = a \tan \theta are used. Visually, these substitutions link the algebraic expression to the geometry of a circle or a right-angled triangle, where the algebraic terms represent the lengths of sides.

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Integration of the Form ∫ex(f(x)+f′(x))dx\int e^x (f(x) + f'(x)) dx: This is a special shortcut result where the integral evaluates directly to exf(x)+Ce^x f(x) + C. This represents a scenario where the total rate of change of a system is the sum of the function and its own derivative, scaled by an exponential growth factor, common in natural growth and decay models.

📐Formulae

∫dxx2−a2=12alog⁡∣x−ax+a∣+C\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \log \left| \frac{x - a}{x + a} \right| + C

∫dxa2−x2=12alog⁡∣a+xa−x∣+C\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a + x}{a - x} \right| + C

∫dxx2+a2=1atan⁡−1(xa)+C\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C

∫dxa2−x2=sin⁡−1(xa)+C\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1} \left( \frac{x}{a} \right) + C

∫dxx2−a2=log⁡∣x+x2−a2∣+C\int \frac{dx}{\sqrt{x^2 - a^2}} = \log \left| x + \sqrt{x^2 - a^2} \right| + C

∫dxx2+a2=log⁡∣x+x2+a2∣+C\int \frac{dx}{\sqrt{x^2 + a^2}} = \log \left| x + \sqrt{x^2 + a^2} \right| + C

∫u⋅v dx=u∫v dx−∫(dudx⋅∫v dx)dx\int u \cdot v \, dx = u \int v \, dx - \int \left( \frac{du}{dx} \cdot \int v \, dx \right) dx

∫ex(f(x)+f′(x))dx=exf(x)+C\int e^x (f(x) + f'(x)) dx = e^x f(x) + C

💡Examples

Problem 1:

Evaluate ∫2x1+x2dx\int \frac{2x}{1 + x^2} dx using the method of substitution.

Solution:

Step 1: Let t=1+x2t = 1 + x^2. Step 2: Differentiate both sides with respect to xx: dtdx=2x\frac{dt}{dx} = 2x, which gives dt=2x dxdt = 2x \, dx. Step 3: Substitute tt and dtdt into the integral: ∫1tdt\int \frac{1}{t} dt. Step 4: Integrate the function: log⁡∣t∣+C\log |t| + C. Step 5: Substitute the value of tt back in terms of xx: log⁡∣1+x2∣+C\log |1 + x^2| + C.

Explanation:

This problem uses substitution because the derivative of the denominator (2x2x) is present in the numerator. By changing the variable, we transform a rational function into a standard reciprocal integral.

Problem 2:

Evaluate ∫xcos⁡x dx\int x \cos x \, dx using integration by parts.

Solution:

Step 1: Identify uu and vv using ILATE. Let u=xu = x (Algebraic) and dv=cos⁡x dxdv = \cos x \, dx (Trigonometric). Step 2: Calculate du=dxdu = dx and v=∫cos⁡x dx=sin⁡xv = \int \cos x \, dx = \sin x. Step 3: Apply the integration by parts formula: ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du. Step 4: Substitute the values: xsin⁡x−∫sin⁡x dxx \sin x - \int \sin x \, dx. Step 5: Integrate the remaining term: xsin⁡x−(−cos⁡x)+Cx \sin x - (-\cos x) + C. Step 6: Simplify: xsin⁡x+cos⁡x+Cx \sin x + \cos x + C.

Explanation:

Integration by parts is applied here because the integrand is a product of an algebraic function and a trigonometric function. Choosing u=xu=x allows the degree of the algebraic part to reduce to 1, making the second integral straightforward.