INTEGRATION BY SUBSTITUTION

Key Idea :

Let us consider the integration of a function with respect to x. If the derivative of a part of the function is at somewhere of the function and you are able to multiply the derivative along with dx using multiplication, then you can integration the function using substitution.

Consider the following integration.

In the function, the power of 'e' is x3. The derivative is x3 is 3x2 and it is being as a part of the function and we will be able to write the 3x2 along with dx using multiplication.

Now substitute a new variable for x3 for which we have the derivative 3x2 being a part of the given function.

Let u = x3.

u = x3

Differentiate with respect to x on both sides.

du/dx = 3x2

Multiply both sides by dx.

du = 3x2dx

In the given integration, substitute x3 = u and 3x2dx = du.

= ∫eudu

= eu + C

Substitute u = x3.

Integrate each of the following with respect to x :

Example 1 :

x/√1 + x2

Solution :

= ∫(x/√1 + x2)dx

= ∫(xdx)/√1 + x2 ----(1)

Let t = 1 + x2.

t = 1 + x2

Differentiate with respect to x.

dt/dx = 0 + 2x

dt/dx = 2x

Multiply both sides by dx.

dt = 2xdx

Divide both sides by 2.

dt/2 = xdx

Substitute 1 + x2 = t and xdx = dt/2 in (1).

∫(xdx)/√1 + x2  = ∫(dt/2)/√t

= (1/2)∫dt/√t

= (1/2)∫dt/t-1/2

  = (1/2)∫t-1/2dt

= (1/2)t-1/2 + 1/(-1/2 + 1) + C

= (1/2)t1/2/(1/2) + C

= (1/2)t1/2(2/1) + C

= t1/2 + C

= √t + C

Substitute t = x2 + 1.

=  √(1 + x2) + C

Example 2 :

x2/(1 + x6)

Solution :

= ∫[x2/(1 + x6)] dx

=  ∫(x2dx)/[1 + (x3)2] ----(1)

Let t = x3.

t = x3

Differentiate with respect to x.

dt/dx = 3x2

Multiply both sides by dx.

dt = 3x2dx

Divide both sides by 3.

dt/3 = x2dx

Substitute x3 = t and x2dx = dt/3 in (1)

∫(x2dx)/(1 + (x3)2 = ∫(dt/3)/(1 + t2)

= (1/3)∫1/(1 + t2)dt

= (1/3)tan-1t + c 

Substitute t = x3.

= (1/3)tan-1(x3) + c

Example 3 :

(ex - e-x)/(ex + e-x)

Solution :

= ∫[(ex - e-x)/(ex + e-x)]dx

= ∫(ex - e-x)dx/(ex + e-x) ----(1)

Let t = ex + e-x.

t = ex + e-x

Differentiate with respect to x.

dt/dx = ex + e-x(-1)

dt/dx = ex - e-x

Multiply both sides by dx.

dt = (ex - e-x)dx

Substitute ex + e-x = t and (ex - e-x)dx = dt in (1).

(ex - e-x)dx/(ex + e-x) = ∫dt/t

= ∫(1/t)dt

= lnt

Substitute t = ex + e-x.

= ln(ex + e-x) + c

Example 4 :

(10x9 + 10xln10)/(10x + x10)

Solution :

= ∫[(10x9 + 10xln10)/(10x + x10)]dx

= ∫(10x9 + 10xln10)dx/(10x + x10) ----(1)

Let u = 10x + x10.

u = 10x + x10

Differentiate with respect to x.

du/dx = 10xln10 + 10x9

Multiply both sides by dx.

du = (10xln10 + 10x9)dx

du = (10x10xln10)dx

Substitute 10x + x10 = u and (10x10xln10)dx in (1).

= ∫du/u

= ∫(1/u)du

= lnu + c

Substitute u = 10x + x10.

= ln(10x + x10) + c

Example 5 :

(sin√x)/√x

Solution :

= ∫[(sin√x)/√x]dx

= ∫(sin√x)(dx/√x) ----(1) 

Let u = √x.

u = √x

u = x1/2

du/dx = (1/2)x1/2 - 1

du/dx = (1/2)x-1/2

du/dx = 1/(2x1/2)

du/dx = 1/(2x)

Multiply both sides by 2dx.

2du = dx/x

Substitute √x = u and dx/x = 2du in (1).

(sin√x)(dx/√x) = (sinu)(2du)

= 2∫sinudu

= 2(-cosu) + c

= -2cosu + c

Substitute u = √x.

= -2cos√x + c

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