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JEE Main 2026 Preparation: Question Papers, Solutions, Mock Tests & Strategy Unacademy » JEE Study Material » Mathematics » Integration of Trigonometric Identities

Integration of Trigonometric Identities

The trigonometric identities can be used to assess several integrals involving trigonometric functions. These allow the integrand to be expressed in a different format, which may make integration easier. An integral can sometimes be assessed using trigonometric substitution. This unit goes through both of these concepts.

Table of Content
  •  

Base Identities 

It’s critical to practice activities to master the concepts described here so that they become second nature.

You should be aware of the significant outcomes by now

coskxdx = 1ksinkx + c

sinkxdx = -1kcoskx + c

We need to be careful when we want to integrate more complex trigonometric functions, such as sin2xdx,sin5x cos4x dx,etc. In some cases, trigonometric identities can be utilised to represent the integrand in a different form that is easier to integrate. An integral can sometimes be calculated by using trigonometric substitution.

Integration of Trigonometric Functions Formulas

The following are the trigonometric identities that we will utilise in this part or that are required to solve questions:

2sinAcosB=sin(A+B)+sin(A-B)

2cosAcosB=cos(A+B)+cos(A-B)

2sinAsinB=cos(A-B)-cos(A+B)

sin2A+cos2A=1

cos2A=cos2A-sin2A=2cos2A-1=1-2sin2A

sin2A=2sinAcosA

1+tan2A=sec2A

Let’s say we’re looking for 0πsin2xdx

The idea is to rewrite the integrand in an alternate form that does not include powers of sinx using a trigonometric identity. One of the ‘double angle’ formulae is the Integration trigonometric identity we’ll utilise here:

cos2A=1-2sin2A

We can write by rearranging this

sin2A=12(1-cos2A)

It’s worth noting that we can use this identity to turn a sin2A expression into one with no powers. As a result, we can write our integral.

0πsin2xdx=0π12(1-cos2x)dx

and this can be assessed in the following way:

0π12(1-cos2x)dx=[12(x-12sin2x)]0π=[12x-14sin2x]0π=π2

Let’s say we’re looking for sin5x cos4x dx

The integrand is a product of the sin5x and cos4x functions. To define the integrand as the sum of two sine functions, we can apply the identity 2sinAcosB=sin(A+B)+sin(A-B)

We have A = 5x and B = 4x.

 sin5x cos4x dx=12(sin9x+sinx)dx=12(-19sin9x-cosx)+c=-118sin9x-12cosx+c

Integrals involving sine and cosine products

We’ll look at integrals of the type sinmxcosnxdx in this section. We can see how to deal with integrals where m is odd in the first example.

Let’s say we’re looking for sin3xcos2xdx

There is no evident identification that will help us here, according to the integrand and the table of identities. However, we’ll rewrite sin3x as sinx sin2x and apply the identity sin2A=1-cos2Ainstead. The rationale for doing so will become clear.

sin3xcos2xdx=(sinx sin2x)cos2xdx=sinx(1-cos2x)cos2xdx

At this point, we can quickly complete the solution by substituting u=cosx, du= -sinxdx:

sinx(1-cos2x)cos2xdx= -(1-u2)u2du=(u4-u2)du=u55-u33+c=15cos5x-13cos3x+c

When m is even and n is odd, we can apply the identity cos2A=1-sin2A and the substitution u=sinx to get the same result.

The use of a trigonometric substitution in integrals

By using a trigonometric substitution, you can find a number of integrals. Consider the following illustration.

Let’s say we’re looking for 11+x2dx

Let us see what happens if we make the change x=tan

The rationale for this is that the integrand will become involved as a result 11+tan2

We also have an identity 1+tan2A=sec2A that allows us to simplify things.

With x=tan, dxd=sec2,

 so that dx=sec2d.

 The integral is transformed into

11+x2dx=11+tan2sec2d=1sec2sec2d=1 d=+c=tan-1x+c

So, 11+x2dx=tan-1x+c.

This is a crucial standard outcome.

We can generalise this result to the integral 1a2+x2dx

We make the necessary changes x=atan, dx=asec2d.The integral is transformed into

1a+atan2asec2d

This is reduced to the below step using the identity 1+tan2A=sec2A

1a21 d=1a+c=1atan-1xa+c

This is an expected outcome that you should be aware of and ready to seek up if necessary.

Let’s say we’re looking for 1a2-x2dx

The replacement we’ll employ is based on the fact that there’s a term as a2-x2 in the denominator and that there’s a  trigonometric identity 1-sin2A=cos2A (and thus a2-a2sin2A=a2cos2A).

We strive to make x=asin, so that x2=a2sin2. Then dxd=acos and  dx=acosd . The integral is transformed into

1a2-x2dx=1a2-a2sin2acosd=1a2cos2acosd=1acosacosd=1d=+c=sin-1xa+c

Hence, 1a2-x2dx=sin-1xa+c

This is yet another typical outcome.

Conclusion

In the preceding paragraph, you’ve examined a number of trigonometric identities. A summary of them is useful for reference. The majority of these identities refer to a single angle denoted, however, there are a few that include two angles, which are designated α and β.

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