Integration of Trigonometric functions involves basic simplification techniques. These techniques use different trigonometric identities which can be written in an alternative form that are more amenable to integration.

The integration of a function f(x) is given by F(x) and it is represented by:

∫f(x)dx = F(x) + C |

Here,

R.H.S. of the equation means integral f(x) with respect to x.

F(x) is called anti-derivative or primitive.

f(x) is called the integrand.

dx is called the integrating agent.

C is called constant of integration or arbitrary constant.

x is the variable of integration.

To understand this concept let us solve some examples.

\(\cos^2 x = \left ( \frac{1 + \cos 2x}{2} \right )\) Form this identity \(2 \cos^2 x = 1 + \cos 2x\) Substituting the above value in the given integrand, we have \(\int 2 \cos^{2}x dx = \int (1+ \cos 2x). dx\) — (1) According to the properties of integration, the integral of sum of two functions is equal to the sum of integrals of the given functions, i.e., \(\int [f(x)+g(x)]dx = \int f(x).dx + \int g(x).dx\) Therefore equation 1 can be rewritten as: \(\int (1 + cos 2x) dx = \int 1 dx + \int cos2xdx\) = \(x + \frac{\sin 2x}{2} + C\) This gives us the required integration of the given function.
\(\sin x \cos y = \frac{1}{2} [\sin (x+y) + \sin (x-y)]\) Form this identity\(\sin 4x \cos 3x = \frac{1}{2} (\sin 7x + \sin x)\) Therefore, \(\int (\sin4x \cos3x)dx = \int \frac{1}{2} (\sin7x + \sin x)dx\) From the above equation we have: \(\int \frac{1}{2}(\sin7x + \sin x)dx = \frac{1}{2} \int (\sin7x + \sin x)dx\) …………(ii) According to the properties of integration, the integral of sum of two functions is equal to the sum of integrals of the given functions, i.e., \(\int [f(x)+g(x)]dx = \int f(x).dx + \int g(x).dx\) Therefore equation 2 can be rewritten as: \(\frac{1}{2} \int (\sin7x ) + \frac{1}{2}\int (\sin x)dx\) = \(\frac{-\cos 7x}{14} + \frac{-\cos x}{2} + C\) This gives us the required integration of the given function.
We know, \(2 \sin x \cos x = \sin 2x\) \(\sin x . \cos x = \frac{\sin 2x}{2}\) Substituting the value in the given integrand, we have \(\int \sin ^{2}x. \cos ^{2}x \; dx = \int (\sin x . \cos x)^{2} dx = \int \left ( \frac{\sin 2x}{2} \right )^{2}\) \(= \frac{1}{4} \int \sin^{2} 2x\) ………(i) Also we know \(\sin^2 x = \frac{1 – \cos 2x }{2}\) Substituting the above value in equation (i), we have \(\frac{1}{4} \int \sin^{2} 2x = \frac{1}{4} \int \frac{1- \cos 4x}{2}\) \(= \int \frac{1}{8} dx – \int \frac{\cos 4x}{8} \; dx\) \(= \frac{1}{8} x + C_{1} – \frac{\sin 4x}{32} + C_{2}\) \(= \frac{1}{8} x – \frac{\sin 4x}{32} + C\) |

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