 # Integration using Trigonometric Identities

In Calculus, we have learned about differential calculus and integral calculus. Differential calculus deals with differentiation, which involves the process of finding the derivative of a function. Whereas integral calculus deals with finding the integral values, which means finding the antiderivative of the function. Thus, integration is the inverse process of differentiation and vice versa. In this article, we will discuss how to solve the integration problems when the integrand involves some trigonometric functions using known identities.

## Integration using Trigonometric Identities Examples

Now, let us discuss the process of solving the integration problems when the integrand has trigonometric functions, such as sine, cosine, tangent, cosecant, secant and cotangent.

Example 1:

Solve: ∫ sin 2x cos 3x dx.

Solution:

Given: ∫ sin 2x cos 3x dx.

Now, by using the trigonometric identity sin x cos y = (½)[sin(x+y) + sin (x-y)]

Therefore, ∫ sin 2x cos 3x dx = (½)[∫ sin 5x dx – ∫sin x dx]

= (½) [(-⅕ cos 5x + cos x] + C

= (-1/10) cos 5x + (½)cos x + C

Therefore, ∫ sin 2x cos 3x dx = (-1/10) cos 5x + (½)cos x + C.

Example 2:

Find the integral of ∫ sin3x dx.

Solution:

Given: ∫ sin3x dx.

∫ sin3x dx = ∫ sin2x sin x dx.

∫ sin3x dx = ∫ (1- cos2x) sin x dx

Now, substitute cos x = t, hence -sinx dx = dt

Therefore,

∫ sin3x dx = -∫ (1-t2) dt

= – ∫dt + ∫t2dt

= -t +(t3/3) + C

As, cos x = t,

∫ sin3x dx = – cos x + (⅓)cos3x + C.

Hence, the integral of ∫ sin3x dx is – cos x + (⅓)cos3x + C.

Example 3:

Solve: ∫sin3x cos2x dx

Solution:

Given: ∫sin3x cos2x dx

∫sin3x cos2x dx = ∫sin2x sin x cos2x dx

We know that sin2x = 1-cos2x.

Therefore,

∫sin3x cos2x dx = ∫(1-cos2x)cos2x sinx dx

Now, put t = cos x, and hence dt = -sin x dx

Hence,

∫sin2x cos2x sin x dx = -∫(1-t2)t2 dt

= -∫(t2-t4) dt

Now, integrate the above function, we get

∫sin2x cos2x sin x dx= -[(t3/3) -(t5/5)] + C

Now, again substitute t = cos x, we get

∫sin2x cos2x sin x dx = -(⅓) cos3x + (⅕)cos5x + C

Therefore, ∫sin3x cos2x dx = -(⅓) cos3x + (⅕)cos5x + C.