Question

How do you find the antiderivative of \[{\cos ^3}(x)dx\] ?

Answer 1

Hint: We first break the power of cosine function and use the trigonometry identity to write \[{\cos ^2}x = 1 - {\sin ^2}x\] . We use the substitution of the sine function as a new variable and convert the complete equation in terms of the new variable.
* Anti-derivative of a function is the opposite of the derivative of a function i.e. it is that value which can be obtained when taking reverse of the derivative function. It is that function whose derivative we take and then we get the function.
* If \[y = {x^n}\] then integration of y with respect to x will be given by the formula: \[\int {ydx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C\] , where C is constant of integration

Complete step by step solution:
We have to solve for the value of \[\int {{{\cos }^3}(x)dx} \] … (1)
We know that power can be broken down using the fact that when the base is the same the powers get added.
We can write \[\int {{{\cos }^3}(x)dx} = \int {\cos x \times \left( {{{\cos }^2}x} \right)dx} \]
Now we know the identity \[{\sin ^2}x + {\cos ^2}x = 1\], which gives us the value \[{\cos ^2}x = 1 - {\sin ^2}x\]
\[ \Rightarrow \int {{{\cos }^3}(x)dx} = \int {\cos x \times \left( {1 - {{\sin }^2}x} \right)dx} \] … (2)
Let us substitute the value of \[\sin x = t\]
Then differentiating both sides of the equation we get
\[ \Rightarrow \cos x dx = dt\]
(as we know differentiation of sine function is cosine function)
Now substitute the value of \[\sin x = t\] and the value of \[\cos x dx = dt\] in equation (2)
\[ \Rightarrow \int {{{\cos }^3}(x)dx} = \int {\left( {1 - {t^2}} \right)dt} \]
Now we know the formula of integration: \[\int {ydx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C\] where C is constant of integration
\[ \Rightarrow \int {{{\cos }^3}(x)dx} = t - \dfrac{{{t^{2 + 1}}}}{{2 + 1}} + C\]
\[ \Rightarrow \int {{{\cos }^3}(x)dx} = t - \dfrac{{{t^3}}}{3} + C\]
Substitute the value of \[\sin x = t\] in the equation
\[ \Rightarrow \int {{{\cos }^3}(x)dx} = \sin x - \dfrac{{{{\sin }^3}x}}{3} + C\]
\[\therefore \]The anti-derivative of \[{\cos ^3}(x)dx\] is \[\sin x - \dfrac{{{{\sin }^3}x}}{3} + C\]

Note: Many students make the mistake of calculating the derivative first and then integrate the obtained derivative which is wrong. Keep in mind we have to calculate the integration of the function, not the derivative. Anti-derivative of a function means that value which on differentiation or derivative gives the function.