
How do you find the antiderivative of $\cos \left( 2x \right)$?
Answer
532.2k+ views
Hint: From the question given we have to find the antiderivative of $\cos \left( 2x \right)$. As we know that antiderivative means indirectly, we have to find the integral of the $\cos \left( 2x \right)$. To find this first we have to assume a variable “u” and it is equal to $2x$. as we know that the integral of $\cos u$ is $\sin u+c$. From this we will get the antiderivative of the $\cos \left( 2x \right)$.
Complete step by step answer:
From the question given we have to find the antiderivative of
$\Rightarrow \cos \left( 2x \right)$
As we know that the antiderivative means indirectly, we have to find the integral, that means we have to find, integration of
$\Rightarrow \int{\cos \left( 2x \right)}$
First, we have to assume a variable,
Let it is “u” and it is equal to $2x$that is,
$\Rightarrow u=2x$
after differentiating on both sides, we will get,
$\Rightarrow du=2dx$
$\Rightarrow \dfrac{1}{2}du=dx$
Now substitute the above in the equation, we will get,
$\Rightarrow \int{\cos \left( u \right)}\dfrac{du}{2}$
As we know that integral of cos is sin, that means,
$\Rightarrow \int{\cos \left( u \right)=\sin \left( u \right)+c}$
By this we will get,
$\Rightarrow \int{\cos \left( u \right)}\dfrac{du}{2}=\dfrac{\sin u}{2}+c$
Since $\Rightarrow u=2x$
By this we will get
$\Rightarrow \int{\cos \left( 2x \right)}=\dfrac{\sin 2x}{2}+c$
Note: Students should know that in integration the chain rule is not applicable so we have to use a substitution method. Students should recall all the formulas of trigonometry and integration, formulas like
$\begin{align}
& \Rightarrow \int{\cos \left( u \right)du=\sin \left( u \right)+c} \\
& \Rightarrow \cos 2x=2{{\cos }^{2}}x-1=1-2{{\sin }^{2}}x={{\cos }^{2}}x-{{\sin }^{2}}x \\
& \Rightarrow \operatorname{Sin}2x=2\operatorname{Sin}x\operatorname{Cos}x \\
& \Rightarrow \int{\sin xdx}=-\cos x+c \\
& \Rightarrow \int{{{\sec }^{2}}xdx}=\tan x+c \\
& \Rightarrow \int{dx}=x+c \\
\end{align}$
Students should not forget to write the plus constant “C” at the end of the solution.
Complete step by step answer:
From the question given we have to find the antiderivative of
$\Rightarrow \cos \left( 2x \right)$
As we know that the antiderivative means indirectly, we have to find the integral, that means we have to find, integration of
$\Rightarrow \int{\cos \left( 2x \right)}$
First, we have to assume a variable,
Let it is “u” and it is equal to $2x$that is,
$\Rightarrow u=2x$
after differentiating on both sides, we will get,
$\Rightarrow du=2dx$
$\Rightarrow \dfrac{1}{2}du=dx$
Now substitute the above in the equation, we will get,
$\Rightarrow \int{\cos \left( u \right)}\dfrac{du}{2}$
As we know that integral of cos is sin, that means,
$\Rightarrow \int{\cos \left( u \right)=\sin \left( u \right)+c}$
By this we will get,
$\Rightarrow \int{\cos \left( u \right)}\dfrac{du}{2}=\dfrac{\sin u}{2}+c$
Since $\Rightarrow u=2x$
By this we will get
$\Rightarrow \int{\cos \left( 2x \right)}=\dfrac{\sin 2x}{2}+c$
Note: Students should know that in integration the chain rule is not applicable so we have to use a substitution method. Students should recall all the formulas of trigonometry and integration, formulas like
$\begin{align}
& \Rightarrow \int{\cos \left( u \right)du=\sin \left( u \right)+c} \\
& \Rightarrow \cos 2x=2{{\cos }^{2}}x-1=1-2{{\sin }^{2}}x={{\cos }^{2}}x-{{\sin }^{2}}x \\
& \Rightarrow \operatorname{Sin}2x=2\operatorname{Sin}x\operatorname{Cos}x \\
& \Rightarrow \int{\sin xdx}=-\cos x+c \\
& \Rightarrow \int{{{\sec }^{2}}xdx}=\tan x+c \\
& \Rightarrow \int{dx}=x+c \\
\end{align}$
Students should not forget to write the plus constant “C” at the end of the solution.
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