Question

Integration by Substitution of Integrals

Answer 1

Integrals of certain functions cannot be obtained directly if they are not in the standard forms, but they may be reduced to standard forms by proper substitution. The method of finding an integral by reducing it to standard form by a proper substitution is called integration by substitution.

Some functions which are integrated by substitution

(i) Integrals of the form If (ax ±b)dx

Let I = ∫f (ax ± b) dx

Putting ax ± b = t

Differentiating both sides with respect to x, we get

adx = dt

or dx = \(\frac{1}{a}\) dt

Substituting ax±b = t and dx = \(\frac{1}{a}\) dt in I, we get

I = ∫f(ax ± b) dx

= \(\frac{1}{a}\) ∫f(t)dt = \(\frac{1}{a}\) F(t) + C

= \(\frac{1}{a}\) F(ax ±b) + C

(ii) Integrals of the form ∫f{Φ(x)} Φ'(x) dx

If Φ(x) is a continuously differentiable function, then we substitute

Φ(x) = t and Φ'(x) dx = dt

This substitution reduces the above integral to ∫(f) dt. After evaluating this integral we substitute back the value of f.

(iii) Integrals of the form ∫ \(\int \frac{\phi (x)}{f \{\phi(x)\}}\) dx

Putting Φ(x) = t

Differentiating both sides with respect to x, we get <|y'(x)dx = dt

Φ'(x)dx = dt

Then, \(\int \frac{\phi (x)}{f \{\phi(x)\}} = \int \frac{dt}{f(t)}\)

After evaluating this integral we substitute back the value of t.

Remark : Method of integration by substitution is also used in other methods of integration like integration by parts and integration by partial function.

Integration by substitution of some important trigonometric functions

Now, we discuss some important integrals involving trigonometric functions and their standard integrals using substitution technique. These will be used later without reference.

See the integrals of the following trigonometric functions with respect to x:

(i) ∫tan x dx = \(\int \frac{sin\ x}{cos\ x}\) dx

Putting cos x = t

Then - sin x dx = dt

or sin x dx = - dt

\(\int \frac{sin\ x}{cos\ x}\) dx = -\(\int \frac{1}{t}\) dt = - log |t| + C

= - log |cos x| + C

Thus, ∫tan x dx = - log |cos x| + C

⇒ ∫tan x dx = log |sec x| + C

(ii) ∫cot x dx = dt

Putting sin x = t

Then, cos x dx = dt

\(\int \frac{cos\ x}{sin\ x}\) dx = -\(\int \frac{1}{t}\) dt = log |t| + C

= log |sin x| + C

Thus, ∫cot xdx = log |sin x| + C

= log |t| + C

= log |sec x + tan x| + C

∴ ∫sec xdx = log |sec x + tan x| + C

Putting cosec x + cot x = t

Then (- cosec x cot x - cosec²x) dx = df

or - (cosec x cot x + cosec²x) dx = dt

or (cosec x cot x + cosec²x) dx = - dt

= - log | t | + C

= - log |cosec x + cot x| + C

= -log

[Multiplying numerator and denominator by (cosec x - cot x)]

= log |cosec x - cot x| + C

∴ ∫cosec x dx = log |cosec x- cotx| +C