Question

How do you integrate \[sin\text{ }3x\text{ }dx\]?

Answer 1

Hint: Integration is a method of adding small quantities to find the whole. Finding an integral is the reverse of finding a derivative.
To solve this question, we are going to use the method of substitution to first substitute \[3x\] as \[u\] and then we can integrate \[\sin u\text{ }\] using the reverse of differential used to get \[sinx\text{ }\]. To the final integral in indefinite integrals, we add a constant $c$ called constant of integration because the derivative of a constant is zero. Integrals of many functions are known and there are other integration rules to find the integral of more complicated functions.

Complete step by step solution:
We have to integrate \[sin\text{ }3x\text{ }dx\]
The symbol of integration is $\int{..}$
So we can represent the given problem as $\int{\sin 3xdx}$
Let us now use substitution
Let \[u=3x\]
Then \[du=3dx\]
This implies that
 \[dx=du/3\]
Therefore we can write $\int{\sin 3xdx}$ as $\int{\sin u\dfrac{du}{3}}$
Now we know that integration of $\sin \theta $ given $-\cos \theta $ because derivative of $\cos \theta $ is $-\sin \theta $
And the constant can directly be taken out of the integration symbol
Therefore $\int{\sin u\dfrac{du}{3}}=\dfrac{1}{3}\int{\sin udu=\dfrac{1}{3}(-\cos u)+c}$ , $c$- constant of integration
Now we can replace all \[u\]'s in the answer by \[3x\]
Thus $\int{\sin 3xdx}=-\dfrac{1}{3}\cos 3x+c$

Therefore, we get integral of \[sin\text{ }3x\text{ }dx\] as $-\dfrac{1}{3}\cos 3x+c$

Note:
We can find out the particular value of the constant of integration if we are doing definite integration. A definite integral has definite values to calculate between. The symbol of integral is actually 'S' which means summing slices and as these slices reach zero, we get the true answer.
Integration is used to find areas, volumes, displacement, etc.