Question

How do you find the antiderivative of \[{\cos ^2}x\]?

Answer 1

Hint:In the given question, we have been given a trigonometric function. We have to find the antiderivative of the trigonometric function. Say \[f\left( x \right)\] is the antiderivative of \[g\left( x \right)\]. Then, it means that when \[f\left( x \right)\] is differentiated, we get \[g\left( x \right)\], or, when we integrate \[g\left( x \right)\], we get \[f\left( x \right)\]. So, we have to find a trigonometric function, which is when differentiated, gives us the given trigonometric function, or simply, we can say that we have to find the result of the integration of the given trigonometric function.

Formula Used:
To find the antiderivative \[\left( {f\left( x \right)} \right)\] of \[g\left( x \right)\], we have:
\[\int {g\left( x \right)} = f\left( x \right)\]

Complete step by step answer:
The given trigonometric function in the question is \[\sin x\]. To find the antiderivative \[\left( {f\left( x \right)} \right)\] of this trigonometric function \[\left( {g\left( x \right)} \right)\], we are going to integrate it,
\[\int {g\left( x \right)} = f\left( x \right)\]
Hence, we have been given \[g\left( x \right)\] which is equal to \[\sin x\] and we have to find the value of \[f\left( x \right)\].
According to the formula,
\[\int {g\left( x \right)} = f\left( x \right)\]
So, putting \[g\left( x \right) = {\cos ^2}x\], which is a standard formula, we get:
\[\int {{{\cos }^2}\left( x \right)dx} = \dfrac{x}{2} + \dfrac{{\sin 2x}}{4} + c\]

Hence, the antiderivative of \[{\cos ^2}\left( x \right)\] is \[\left( {\dfrac{x}{2} + \dfrac{{\sin 2x}}{4} + c} \right)\].

Note: This can be calculated only by learning the integrals of the trigonometric functions. There is no method to calculate them. So, it is necessary that we learn the integrals of all trigonometric functions. We can see that the calculated result is the correct antiderivative of the given function, as if we find the derivative of the result, we get back the function which is given in the question, hence, the meaning of ‘antiderivative’.