Question

How do you find the antiderivative of $1-\cos \left( 4x \right)$?

Answer 1

Assume the value of the given integral as ‘I’. Break the integral into two parts and integrate each of the terms: - 1 and $\cos \left( 4x \right)$. Use the basic integral formula for the integration of the cosine function given as $\int{\left( \cos \left( ax+b \right) \right)}dx=\dfrac{1}{a}\sin \left( ax+b \right)$. Here, ‘a’ and ‘b’ are real constants. To integrate the constant term 1, write it as ${{x}^{0}}$ and then evaluate using the formula $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$. Add the constant of indefinite integration ‘C’ at last to get the answer.

Complete step by step solution:
Here, we have been provided with the function $1-\cos \left( 4x \right)$ and we are asked to find its antiderivative, in other words we have to integrate this function. Let us assume the integral as I, so we have,
$\Rightarrow I=\int{\left( 1-\cos \left( 4x \right) \right)dx}$
Breaking the integral into two parts, one for each term, we have,
$\Rightarrow I=\int{1dx}-\int{\cos \left( 4x \right)dx}$
Now, we can write the constant term 1 as ${{x}^{0}}$, so we have,
$\Rightarrow I=\int{{{x}^{0}}dx}-\int{\cos \left( 4x \right)dx}$
Now, applying the basic formula of integral given as: - $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$, we get,

& \Rightarrow I=\dfrac{{{x}^{0+1}}}{0+1}-\int{\cos \left( 4x \right)dx} \\\ & \Rightarrow I=x-\int{\cos \left( 4x \right)dx} \\\ \end{aligned}$$ Now, we need to find the integral of the cosine function. Using the formula: $\int{\left( \cos \left( ax+b \right) \right)}dx=\dfrac{1}{a}\sin \left( ax+b \right)$, where ‘a’ and ‘b’ are real constants, we get, $$\Rightarrow I=x-\dfrac{1}{4}\sin \left( 4x \right)$$ Now, since the given integral was an indefinite integral and therefore we need to add a constant of integration (C) in the expression obtained for I. So, we get, $$\Rightarrow I=x-\dfrac{1}{4}\sin \left( 4x \right)+C$$ Hence, the above relation is required answer **Note:** One may note that the formula: $$\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$$ is invalid for n = -1. This is because in this case (n + 1) will become 0 and the integral will become undefined. So, when n = -1 then the function becomes $$\int{\dfrac{1}{x}dx}$$ whose solution is $$\ln x$$. In the above question do not use the formula: $1-\cos 2\theta =2{{\sin }^{2}}\theta $ and find the integral, this is because we do not have any direct integral formula for the function ${{\sin }^{2}}\theta $. At last, do not forget to add the constant of integration (C) as we are finding indefinite integral and not definite integral.