
What is the opposite of derivative in calculus?
Answer
505.2k+ views
Hint: For solving this question you should know about finding integration of any trigonometric function. In this question it is asked to determine the antiderivative, which means integration of function and we will divide our function into two parts and then we will solve that and get our answer. Let’s see an example for this. Find the antiderivative of ${{\left( \sin x \right)}^{3}}$. According to the question we have to find the antiderivative or integration of ${{\left( \sin x \right)}^{3}}$.
Complete step-by-step solution:
As we know, the antiderivative of any trigonometric function will be equal to the integration of the same function. And we can understand it by an example.
Example 1: Find the derivative of $y={{x}^{3}}$ and also find the antiderivative of the answer for this.
For the derivative of ${{x}^{3}}$, we differentiate it with respect to $x$:
$\begin{align}
& \Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{x}^{3}} \right) \\
& \Rightarrow \dfrac{dy}{dx}=3{{x}^{2}} \\
\end{align}$
Now take the antiderivative of $3{{x}^{2}}$.
It means that we find the integration of $3{{x}^{2}}$.
So, the antiderivative of,
$\begin{align}
& 3{{x}^{2}}=\int{3.{{x}^{2}}dx} \\
& =3\dfrac{{{x}^{3}}}{3}+C \\
& ={{x}^{3}} \\
\end{align}$
So, it is equal to our function and it is proved that the antiderivative is the integration of that term.
So, according to the question:
$\begin{align}
& \int{{{\left( \sin x \right)}^{3}}dx} \\
& \Rightarrow \int{{{\left( \sin x \right)}^{3}}dx}=\int{{{\sin }^{2}}x}\sin x.dx \\
& =\int{\left( 1-{{\cos }^{2}}x \right)\sin xdx} \\
& =\int{\sin x.dx+\int{{{\cos }^{2}}x\left( -\sin x \right)dx}} \\
& =-\cos x+\dfrac{{{\cos }^{3}}x}{3}+C \\
\end{align}$
If $\cos x=u$, then $du=\left( -\sin x \right).dx$.
So, the antiderivative of ${{\left( \sin x \right)}^{3}}$ is $-\cos x+\dfrac{{{\cos }^{3}}x}{3}+C$.
Note: While calculating the antiderivative or integration of any trigonometric function you should always use the trigonometric formulas for integration. And make sure that all the calculations and specially ensure the correct powers while changing when we have to integrate them.
Complete step-by-step solution:
As we know, the antiderivative of any trigonometric function will be equal to the integration of the same function. And we can understand it by an example.
Example 1: Find the derivative of $y={{x}^{3}}$ and also find the antiderivative of the answer for this.
For the derivative of ${{x}^{3}}$, we differentiate it with respect to $x$:
$\begin{align}
& \Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{x}^{3}} \right) \\
& \Rightarrow \dfrac{dy}{dx}=3{{x}^{2}} \\
\end{align}$
Now take the antiderivative of $3{{x}^{2}}$.
It means that we find the integration of $3{{x}^{2}}$.
So, the antiderivative of,
$\begin{align}
& 3{{x}^{2}}=\int{3.{{x}^{2}}dx} \\
& =3\dfrac{{{x}^{3}}}{3}+C \\
& ={{x}^{3}} \\
\end{align}$
So, it is equal to our function and it is proved that the antiderivative is the integration of that term.
So, according to the question:
$\begin{align}
& \int{{{\left( \sin x \right)}^{3}}dx} \\
& \Rightarrow \int{{{\left( \sin x \right)}^{3}}dx}=\int{{{\sin }^{2}}x}\sin x.dx \\
& =\int{\left( 1-{{\cos }^{2}}x \right)\sin xdx} \\
& =\int{\sin x.dx+\int{{{\cos }^{2}}x\left( -\sin x \right)dx}} \\
& =-\cos x+\dfrac{{{\cos }^{3}}x}{3}+C \\
\end{align}$
If $\cos x=u$, then $du=\left( -\sin x \right).dx$.
So, the antiderivative of ${{\left( \sin x \right)}^{3}}$ is $-\cos x+\dfrac{{{\cos }^{3}}x}{3}+C$.
Note: While calculating the antiderivative or integration of any trigonometric function you should always use the trigonometric formulas for integration. And make sure that all the calculations and specially ensure the correct powers while changing when we have to integrate them.
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