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.

Video Lesson on Trigonometry

Practice Problems

Find the integral of the following functions using trigonometric identities.

1. Sin^-1 (cos x)
2. Tan⁴x
3. (cos x – sin x)/ (1+ sin 2x)

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