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Integrals - Methods of Integration

Grade 12CBSE

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

🔑Concepts

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Integration by Substitution: Used when the integrand contains a function f(x)f(x) and its derivative f′(x)f'(x). We substitute u=f(x)u = f(x) such that du=f′(x)dxdu = f'(x)dx to simplify the integral.

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Integration by Partial Fractions: Used for integrating rational functions P(x)Q(x)\frac{P(x)}{Q(x)}. If the degree of P(x)<Q(x)P(x) < Q(x), we decompose it into simpler fractions based on the factors of Q(x)Q(x).

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Integration by Parts: Based on the product rule of differentiation. For two functions uu and vv, the formula is ∫uvdx=u∫vdx−∫(u′∫vdx)dx\int u v dx = u \int v dx - \int (u' \int v dx) dx. The choice of uu is usually guided by the ILATE rule (Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential).

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Integrals of Special Functions: Specific formulae are applied for integrands involving quadratic expressions in the denominator, such as 1ax2+bx+c\frac{1}{ax^2+bx+c} or 1ax2+bx+c\frac{1}{\sqrt{ax^2+bx+c}} by completing the square.

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Integration using Trigonometric Identities: Transforming powers of trigonometric functions into multiple angles using identities like cos⁡2x=1+cos⁡2x2\cos^2 x = \frac{1 + \cos 2x}{2} and sin⁡2x=1−cos⁡2x2\sin^2 x = \frac{1 - \cos 2x}{2}.

📐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

∫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

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

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

∫a2−x2dx=x2a2−x2+a22sin⁡−1xa+C\int \sqrt{a^2 - x^2} dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C

💡Examples

Problem 1:

Evaluate ∫2x1+x2dx\int \frac{2x}{1+x^2} dx.

Solution:

Let 1+x2=t1+x^2 = t. Differentiating both sides, we get 2xdx=dt2x dx = dt. Substituting these into the integral: ∫1tdt=log⁡∣t∣+C\int \frac{1}{t} dt = \log |t| + C Substituting back t=1+x2t = 1+x^2, we get log⁡(1+x2)+C\log (1+x^2) + C.

Explanation:

This is a direct application of the Integration by Substitution method because the numerator is the derivative of the denominator.

Problem 2:

Evaluate ∫xexdx\int x e^x dx using Integration by Parts.

Solution:

Using ∫uvdx=u∫vdx−∫(u′∫vdx)dx\int u v dx = u \int v dx - \int (u' \int v dx) dx: ∫xexdx=x∫exdx−∫(ddx(x)∫exdx)dx\int x e^x dx = x \int e^x dx - \int (\frac{d}{dx}(x) \int e^x dx) dx =xex−∫(1⋅ex)dx=xex−ex+C=ex(x−1)+C= x e^x - \int (1 \cdot e^x) dx = x e^x - e^x + C = e^x(x-1) + C

Explanation:

We use the ILATE rule to choose uu. Here, xx is algebraic (A) and exe^x is exponential (E). So, let u=xu = x and v=exv = e^x.

Problem 3:

Evaluate ∫dx(x+1)(x+2)\int \frac{dx}{(x+1)(x+2)}.

Solution:

Let 1(x+1)(x+2)=Ax+1+Bx+2\frac{1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}. Multiplying by (x+1)(x+2)(x+1)(x+2), we get 1=A(x+2)+B(x+1)1 = A(x+2) + B(x+1). Putting x=−1x = -1, 1=A(1)  ⟹  A=11 = A(1) \implies A = 1. Putting x=−2x = -2, 1=B(−1)  ⟹  B=−11 = B(-1) \implies B = -1. ∫(1x+1−1x+2)dx=log⁡∣x+1∣−log⁡∣x+2∣+C=log⁡∣x+1x+2∣+C\int (\frac{1}{x+1} - \frac{1}{x+2}) dx = \log |x+1| - \log |x+2| + C = \log \left| \frac{x+1}{x+2} \right| + C

Explanation:

We use the method of Partial Fractions to split the integrand.