✅ Integration by Substitution Formula ⭐️⭐️⭐️⭐️⭐

Integration by Substitutions

In order to find integrals of functions effectively, we need to develop techniques that can reduce the functions to standard forms. The substitution method is one such technique which we will discuss in detail in this article.

The Substitution Method

According to the substitution method, a given integral ∫ f(x) dx can be transformed into another form by changing the independent variable x to t. This is done by substituting x = g (t).

Consider, I = ∫ f(x) dx
Now, substitute x = g(t) so that, dx/dt = g’(t) or dx = g’(t)dt.
Therefore, I = ∫ f(x) dx = ∫ f[g(t)] g’(t)dt

It is important to note here that you should make the substitution for a function whose derivative also occurs in the integrand as shown in the following examples.

Example 1

Integrate sin(mx) with respect to x.

Solution: We know that the derivative of mx = m. Hence, let’s substitute mx = t, so that mdx = dt. Therefore,

∫ sin mx dx = 1/m ∫ sin t dt
= – 1/m cos t + C
= – 1/m cos mx + C

Example 2

Integrate 2x sin (x² + 1) with respect to x.

Solution: We know that the derivative of (x² + 1) = 2x. Hence, let’s substitute (x² + 1) = t, so that 2x dx = dt. Therefore,

∫ 2x sin (x² + 1) dx = ∫ sin t dt
= – cos t + C
= – cos (x² + 1) + C

Example 3

Integrate {(tan⁴ √x) (sec² √x)} / √x with respect to x.

Solution: We know that the derivative of √x = ½ x^–½. Hence, let’s substitute √x = t, so that ½ (√x) dx = dt or, dx = 2t dt. Therefore,

∫ {(tan⁴ √x) (sec² √x)} / √x = ∫ {(2t tan⁴ t) (sec² t) dt} / t
= 2 ∫ (tan⁴ t) (sec² t) dt

Now, let’s substitute tan t = u, so that sec² t dt = du. Therefore,

2 ∫ (tan⁴ t) (sec² t) dt = 2 ∫ u⁴ du
= 2 (u⁵/5) + C
= (2/5) tan⁵ t + C
= (2/5) tan⁵ √x + C

Hence, ∫ {(tan⁴ √x) (sec² √x)} / √x = (2/5) tan⁵ √x + C

Substitution Method for Some Important Integrals of Trigonometric Functions

• ∫ tan x dx = log |sec x| + C

We know that tan x = sin x / cos x. Therefore,
∫ tan x dx = ∫ (sin x / cos x) dx.
Now, let’s substitute cos x = t, so that sin x dx = – dt. Therefore,
∫ tan x dx = – ∫ (dt / t) = – log |cos x| + C
Or, ∫ tan x dx = log |sec x| + C

• ∫ cot x dx = log |sin x| + C

We know that cot x = cos x / sin x. Therefore,
∫ cot x dx = ∫ (cos x / sin x) dx.
Now, let’s substitute sin x = t, so that cos x dx = – dt. Therefore,
∫ cot x dx = ∫ (dt / t) = log |t| + C = log |sin x| + C

• ∫ sec x dx = log |sec x + tan x| + C

On multiplying both the numerator and denominator by (sec x + tan x), we have
∫ sec x dx = ∫ {sec x (sec x + tan x) dx} / (sec x + tan x)
Now, let’s substitute (sec x + tan x) = t, so that sec x (sec x + tan x) dx = dt.
Therefore, ∫ sec x dx = ∫ (dt / t) = log |t| + C = log |sec x + tan x| + C

• ∫ cosec x dx = log |cosec x – cot x| + C

On multiplying both the numerator and denominator by (cosec x + cot x), we have
∫ cosec x dx = ∫ {cosec x (cosec x + cot x) dx} / (cosec x + cot x)
Now, let’s substitute (x + cot x) = t, so that – cosec x (cosec x + cot x) dx = dt.
Therefore, ∫ cosec x dx = – ∫ (dt / t) = – log |t| + C = – log |cosec x + cot x| + C
= – log |(cosec² x – cot² x) / (cosec x – cot x)| + C
= log |cosec x – cot x| + C

Example 4

Find the integral of (sin³ x) (cos² x) dx

Solution: We have,
∫ (sin³ x) (cos² x) dx = ∫ (sin² x) (cos² x) (sin x) dx

We know that sin² x = (1 – cos² x). Replacing it above, we get
∫ (1 – cos² x) (cos² x) (sin x) dx

Now, let’s substitute cos x = t, so that – sin x dx = dt. Therefore,
∫ (1 – cos² x) (cos² x) (sin x) dx = – ∫ (1 – t²) t² dt
= – ∫ (t² – t⁴) dt
= – [(t³/3) – (t⁵/5)] + C
Hence, ∫ (sin³ x) (cos² x) dx = – (1/3) cos³ x + (1/5) cos⁵ x + C

Example 5

Find the integral of [sin x / sin (x + a)] dx.

Solution: Let’s substitute (x + a) = t, so that dx = dt. Therefore,
∫ [sin x / sin (x + a)] dx = ∫ [sin (t – a) / sin t] dt
= ∫ (sin t cos a – cos t sin a) / sin t] dt
= cos a ∫ dt – sin a ∫ cot t dt = cos a (t) – (sin a) [log |sin t| + C₁] = cos a (x + a) – (sin a) [log |sin (x + a)| + C₁]
= x cos a + a cos a – (sin a) [log sin (x + a)] – C₁ sin a
Hence, ∫ [sin x / sin (x + a)] dx = x cos a – sin a log |sin (x + a)| + C
Where, C = – C₁ sin a + a cos a … an arbitrary constant.

Remember: When the integrand involves some trigonometric functions, we use some known identities to find the integral as illustrated in the following example:

Example 6

Find ∫ cos² x dx

Solution: Recall the identity, cos 2x = 2 cos² x – 1. We can hence derive, cos² x = (1 + cos 2x) / 2. Therefore,
∫ cos² x dx = ½ ∫ (1 + cos 2x) dx = ½ ∫ dx + ½ ∫ cos 2x dx
= x/2 + (¼) sin 2x + C

More Solved Examples for You

Question 1: Find the integral of sin² (2x + 5).

Answer : We know that, cos 2t = 1 – 2 sin² t. Hence, sin² t = ½ (1 – cos 2t). Therefore,
∫ sin² (2x + 5) dx = ½ ∫ [1 – cos 2(2x + 5)] dx = ½ ∫ 1 – cos (4x + 10) dx
= ½ ∫ 1 dx – ½ ∫ cos (4x + 10) dx

We also know that,
∫ cos (ax + b) dx = {sin (ax + b) / a} + C

Hence, ∫ sin² (2x + 5) dx = x/2 – ½ {[sin (4x + 10) / 4] + C}
= x/2 – 1/8 sin (4x + 10) + C

Question 2: Define the substitution method?

Answer: It is a way to simplify the system of equations by expressing one variable in terms of another, therefore removing one variable from an equation. After that, solve this equation and then back substitute until the solution is found.

Question 3: Why we use the substitution method?

Answer: The objective of the substitution method is to rewrite one of the equations in terms of a single variable. Moreover, the important thing here is that you are always substituting values that are equivalent.

Question 4: What are the steps for the substitution method?

Answer: For solving systems of equations follow these steps:

• Firstly, solve one equation for 1 variable (such as put value in y = or x = form).
• Next, substitute this expression into the other equation and solve for the missing variable.
• After that, substitute your answer into the first equation then solve it to get an answer.

Question 5: How to use the substitution method for two variables?

Answer: Solving for two variables, follow the procedure:

• First of all, choose one equation and solve for one of its variables.
• After that, substitute the variable you just solve in the other equation.
• Now, solve the new equation.
• Lastly, substitute the value found into any equation and solves for the other variable.

Integration by Substitution

Integration by substitution is an important method of integration, which is used when a function to be integrated, is either a complex function or if the direct integration of the function is not feasible. The integral of a function is simplified by this method of integration by substitution, by reducing the given function into a simplified function.

Let us learn the process of integration by substitutions, check some of the important substitutions, and also check the solved examples.

What Is Integration by Substitution?

Integration by substitution is used when the integration of the given function cannot be obtained directly, as the given algebraic function is not in the standard form. Further, the given function can be reduced to the standard form by appropriate substitution. Let us consider the indefinite integral of a function f(x), ∫f(x).dx

. Here this integral can be transformed to another form by replacing x with g(t) and by substituting x = g(t).

I = ∫f(x).dx

x = g(t) such that dx/dt = g'(t)

dx = g'(t).dt

I =∫f(x).dx=∫f(g(t)).g′(t).dt

Steps to Integration by Substitution

The following are the steps that are helpful in performing this method of integration by substitution.

Step – 1: Choose a new variable t for the given function to be reduced.

Step – 2: Determine the value of dx, of the given integral, where f(x) is integrated with respect to x.

Step – 3: Make the required substitution in the function f(x), and the new value dx.

Step – 4: Integrate the function obtained after substitution

Step – 5: Substitute back the initial variable x to obtain the final answer.

Important Substituions in Integration by Substitution

The following are some of the important substitutions which are helpful in simplifying the given expression and easily performing the integration process. Let us check the following specific substitutions for integration by substitution.

Examples on Integration by Substitution

So ∫cos(x²) 2x dx = sin(x²) + C

That worked out really nicely! (Well, I knew it would.)

But this method only works on some integrals of course, and it may need rearranging:

Now let’s try a slightly harder example:

Now put u=x²+1 back again:

½ ln(x²+1) + C

And how about this one:

Now get some practice, OK?

In Summary

When we can put an integral in this form:

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