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Methods of Integration> Integration by Substitution

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Estimated time: 24 minutes
CBSE: Class 12
Maharashtra State Board: Class 12

Definition: Integration by Substitution

Integration by substitution is a method in which we replace a part of the integral by a new variable to simplify the integration.

General Formula:

If \[x = g(t), \ dx = g'(t) dt\] then \[\int f(x) dx = \int f(g(t))g'(t) dt\]

CBSE: Class 12
Maharashtra State Board: Class 12

Standard Substitution

Sr. No. Integrand Form Substitution
i \[\sqrt{\mathrm{a}^2-x^2},\frac{1}{\sqrt{\mathrm{a}^2-x^2}},\mathrm{a}^2-x^2\] x = a sinθ or x = a cosθ
ii \[\sqrt{x^2+\mathrm{a}^2},\frac{1}{\sqrt{x^2+\mathrm{a}^2}},x^2+\mathrm{a}^2\] x = a tanθ
iii \[\sqrt{x^{2}-a^{2}},\frac{1}{\sqrt{x^{2}-a^{2}}},x^{2}-a^{2}\] x = a secθ
iv \[\sqrt{\frac{x}{a+x}},\sqrt{\frac{a+x}{x}},\]\[\sqrt{x(a+x)},\frac{1}{\sqrt{x(a+x)}}\] x = a tan²θ
v \[\sqrt{\frac{x}{a-x}},\sqrt{\frac{a-x}{x}},\]\[\sqrt{x(a-x)},\frac{1}{\sqrt{x(a-x)}}\] x = a sin²θ
vi \[\sqrt{\frac{x}{x-a}},\sqrt{\frac{x-a}{x}},\]\[\sqrt{x(x-\mathrm{a})},\frac{1}{\sqrt{x(x-\mathrm{a})}}\] x = a sec²θ
vii \[\sqrt{\frac{\mathrm{a}-x}{\mathrm{a}+x}},\sqrt{\frac{\mathrm{a}+x}{\mathrm{a}-x}}\] x = a cos 2θ
viii \[\sqrt{\frac{x-\alpha}{\beta-x}},\sqrt{(x-\alpha)(\beta-x)},\]\[(\beta>\alpha)\] x = α cos²θ + β sin²θ
CBSE: Class 12

Stepwise Method

  1. Identify the complicated part of the integrand.

  2. Choose a substitution such as \(u = g(x)\).

  3. Differentiate to get \(du\) and rewrite \(dx\) accordingly.

  4. Express the entire integral in terms of the new variable.

  5. Integrate in the simpler form.

  6. Replace the new variable by the original expression.

  7. Add the constant of integration for indefinite integrals.

Example 1

Find the following integrals:

  1. \[\int \sin^3 x \cos^2 x dx\] 
  2. \[\int \frac{\sin x}{\sin (x + a)} dx\]

Solution:

(i) We have

\[\int \sin^3 x \cos^2 x dx = \int \sin^2 x \cos^2 x (\sin x) dx\]

\[= \int (1 - \cos^2 x) \cos^2 x (\sin x) dx\]

Put \[t = \cos x\] so that \[dt = -\sin x dx\]

Therefore,  \[\int \sin^2 x \cos^2 x (\sin x) dx = -\int (1 - t^2) t^2 dt\]

\[= -\int (t^2 - t^4) dt = -\left( \frac{t^3}{3} - \frac{t^5}{5} \right) + \text{C}\]

\[= -\frac{1}{3} \cos^3 x + \frac{1}{5} \cos^5 x + \text{C}\]

(ii) Put \[x + a = t\]. Then \[dx = dt\].

Therefore

\[\int \frac{\sin x}{\sin (x + a)} dx = \int \frac{\sin (t - a)}{\sin t} dt\]

\[= \int \frac{\sin t \cos a - \cos t \sin a}{\sin t} dta \]

\[= \cos a \int dt - \sin a \int \cot t dta\]

\[= (\cos a) t - (\sin a) [\log |\sin t| + \text{C}_1]\]

\[= (\cos a) (x + a) - (\sin a) [\log |\sin (x + a)| + \text{C}_1][= x \cos a + a \cos a - (\sin a) \log |\sin (x + a)| - \text{C}_1 \sin a\]

Hence, a \[\int \frac{\sin x}{\sin (x + a)} dx = x \cos a - \sin a \log |\sin (x + a)| + \text{C}\],

where, \[\text{C} = -\text{C}_1 \sin a + a \cos a\], is another arbitrary constant.

CBSE: Class 12
Maharashtra State Board: Class 12

Key Points: Standard Substitution

  • Integration by substitution is the reverse process of the chain rule.

  • Choose the substitution so that the integral becomes simpler, not more complicated.

  • Always rewrite both the function and \(dx\) in terms of the new variable.

  • For indefinite integrals, back-substitute and add \(C\).

  • For definite integrals, limits should also be changed if the solution is continued in the new variable.

  • Trigonometric substitution is mainly used for radicals involving \(a^2-x^2\), \(x^2+a^2\), and \(x^2-a^2\).

Video Tutorials

We have provided more than 1 series of video tutorials for some topics to help you get a better understanding of the topic.

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Shaalaa.com | Indefinite Integration by U Substitution

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Indefinite Integration by U Substitution [00:29:56]
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