Question

Integrate $ \cos ecx\;dx $

Answer 1

Hint: In mathematics, integration is the concept of calculus and it is the act of finding the integrals. Here we will find integration, by using the concept of equivalent value, where the same term will be multiplied and divided and will simplify and then will place the formula and simplify for the resultant answer. We take use of the method of substitution here.

Complete step-by-step answer:
Take the given function –
 $ I = \int {\cos ecx} \;dx $
Remember when you multiply and divide the term with the same term value remains the same. So, multiply and divide with $ (\cos ecx - \cot x) $ to the given function.
 $ \Rightarrow I = \int {\cos ecx} \; \times \dfrac{{(\cos ecx - \cot x)}}{{(\cos ecx - \cot x)}}dx $ .... (A)
Let us assume that $ (\cos ecx - \cot x) = t $ .... (B)
Take differentiation on both the side-
 $ \Rightarrow ( - \cos ecx\cot x + \cos e{c^2}x)dx = dt $
Make “dx” as the subject. When the term multiplicative on one side is moved to the opposite side, then it goes to division.
 $ \Rightarrow dx = \dfrac{{dt}}{{( - \cos ecx\cot x + \cos e{c^2}x)}} $
Place the above value in the equation (A)
 $ \Rightarrow I = \int {\cos ecx} \; \times \dfrac{{(\cos ecx - \cot x)}}{{(\cos ecx - \cot x)}} \times \dfrac{{dt}}{{( - \cos ecx\cot x + \cos e{c^2}x)}} $
Take a common multiple in the denominator.
 $ \Rightarrow I = \int {\cos ecx} \; \times \dfrac{{(\cos ecx - \cot x)}}{{(\cos ecx - \cot x)}} \times \dfrac{{dt}}{{\cos ecx( - \cot x + \cos ecx)}} $
The above equation can be re-written as -
 $ \Rightarrow I = \int {\cos ecx} \; \times \dfrac{{(\cos ecx - \cot x)}}{{(\cos ecx - \cot x)}} \times \dfrac{{dt}}{{\cos ecx(\cos ecx - \cot x)}} $
Common factors from the numerator and the denominator cancel each other.
 \[ \Rightarrow I = \int {} \dfrac{{dt}}{{(\cos ecx - \cot x)}}\]
From equation (B)
 \[ \Rightarrow I = \int {} \dfrac{{dt}}{t}\]
Apply the integration formula-
 \[ \Rightarrow I = \log \left| t \right| + c\]
Again, by using equation (B)
 \[ \Rightarrow \int {\cos ecx} = \log \left| {\cos ecx - \cot x} \right| + c\] is the required solution.
So, the correct answer is “\[\log \left| {\cos ecx - \cot x} \right| + c\]”.

Note: Remember the difference between the differentiation and the integration and apply the formula accordingly. Differentiation can be represented as the rate of change of the function, whereas integration represents the sum of the function over the range. They are inverse of each other.