Trigonometric functions Integrals: If u is a differentiable function of x , then sin u is a differentiable of x . The chain rule gives the derivative of sin u as d/dx sin u = cos u du/dx . From another point of view , however , this same equation says that sin u is one of the antiderivatives of the product cos u .(du/dx) .
Therefore , integration (cos u du/dx ) dx = sin u + C .
A formal cancellation of the dx’s in the integral on the left leads to the following rule . If u is a differentiable function , then integration cos u du = Sin u + c .
Let us take an example of Trigonometric Integral top understand the concept . suppose vwe have given to integrate cos ( 7 (theta) +5) d(theta) so , to solve this let us take u = 7 (theta) + 5 , du = 7 d(theta) => 1/7 du = d(theta) so plugging in the problem we have integration cos u . 1/7 du = 1/7 integration cos u du = 1/7 sin u + c .
The chain rule formulae for the derivatives of the tangent , cotangent , secant , and cosecant of a differentiable function will give us the following Trigonometric Integrals Table:
Table of Trigonometric Integrals
(i) Integration of sin x dx = cos x + c , Extension sin( ax +b) dx = -cos (ax+b) / a+ c
(ii) Integration of cos x dx = sin x + c , extension integration cos (ax + b) = sin (ax +b) / a+ c
(iii) Integration tan x dx = -log mod (cos x ) + c , extension integration tan ( ax+ b) = -log mod ( cos (ax +b)/ a+ c
(iv) Integration cot x dx = - cosec ^2 x + c , extension integration cot (ax +b) = -cosec^2(ax +b) / a+ c
(v) Integration cosec x dx = log mod (cosec x - cot x ) + c , extension integration cosec (ax +b) dx = - log mod( cosec (ax +b) – cot (ax +b))/ a+ c
(vi) Integration sec x dx = log mod (sec x + tanx ) + c , extension sec (ax + b) dx = log mod sec (ax +b) + tan (ax +b)/a + c
(vii) Integration sec x . tan x = sec x + c , extension integration sec (ax + b).tan (ax+b) dx = sec (ax +b)/ a+ c
(viii) Integration cosec x . cot x dx –cosec x + c , extension cosec (ax + b) . cot (ax + b) x = -cosec (ax +b) / a+ c
