 # 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.

### Suggested Videos        ## 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. (Source: flickr)

### 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?

• 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?

• 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.

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