Question

Solve the following $\int{\sin 2x\cos 3x}dx$

Answer 1

Hint: We need to find the integral of the function $\sin 2x\cos 3x$ . We start to solve the question by multiplying and dividing the integral by 2. Then, we use the trigonometric formula $2\sin A\cos B=\sin \left( A+B \right)+\sin \left( A-B \right)$ to simplify the trigonometric function and integrate it to get the desired result.

Complete step by step solution:
Let $I$ be the value of the integral for the given function.
$\Rightarrow I=\int{\sin 2x\cos 3x}dx$
We are given a function and need to integrate it. We solve this question using the trigonometric formulae to simplify the function and then find the value of $I$ .
According to the question,
The integral of the function $\sin 2x\cos 3x$ is written as follows,
$\Rightarrow I=\int{\sin 2x\cos 3x}dx$
We need to multiply and divide by 2.
Multiplying and dividing by 2 on the right-hand side of the equation, we get,
$\Rightarrow I=\dfrac{1}{2}\times 2\int{\sin 2x\cos 3x}dx$
$\Rightarrow I=\dfrac{1}{2}\int{2\sin 2x\cos 3x}dx$
The above trigonometric function is of the form $2\sin A\cos B$
From trigonometry,
We know that $2\sin A\cos B=\sin \left( A+B \right)+\sin \left( A-B \right)$ .
Here,
The values of $A$ and $B$ are given as follows,
$A=2x;$
$B=3x$
Applying the above formula and substituting the values in the formula, we get,
$\Rightarrow I=\dfrac{1}{2}\int{\left( \sin \left( 2x+3x \right)+\sin \left( 2x-3x \right) \right)dx}$
Simplifying the value of the above equation, we get,
$\Rightarrow I=\dfrac{1}{2}\int{\left( \sin 5x+\sin \left( -x \right) \right)dx}$
From trigonometry,
We know that $\sin x$ is an odd function.
For any odd function,
$\Rightarrow f\left( x \right)=-f\left( x \right)$
Applying the same for the $\sin x$ function, we get,
 $\Rightarrow \sin \left( -x \right)=-\sin x$
Substituting the same, we get,
$\Rightarrow I=\dfrac{1}{2}\int{\left( \sin 5x-\sin x \right)dx}$
Let us evaluate the above equation further,
$\Rightarrow I=\dfrac{1}{2}\int{\sin 5xdx}-\dfrac{1}{2}\int{\sin xdx}$
From the formulae of integration,
$\Rightarrow \int{\sin 5xdx=\dfrac{\left( -\cos 5x \right)}{5}}$
$\Rightarrow \int{\sin xdx=\left( -\cos x \right)}$
Substituting the values of integrals in the above equation, we get,
$\Rightarrow I=\dfrac{1}{2}\dfrac{\left( -\cos 5x \right)}{5}-\dfrac{1}{2}\left( -\cos x \right)$
Simplifying the above equation, we get
$\Rightarrow I=\dfrac{\left( -\cos 5x \right)}{10}+\dfrac{\cos x}{2}$
$\Rightarrow I=\dfrac{1}{2}\cos x-\dfrac{1}{10}\cos 5x$
Substituting the value of $I$ in the above equation, we get,
$\therefore \int{\sin 2x\cos 3xdx}=\dfrac{1}{2}\cos x-\dfrac{1}{10}\cos 5x$

Note: One must always remember that the difference between the trigonometric functions $\sin 5x-\sin x$ is not equal to $\sin 4x$ and it is solved using the formula $\sin \left( A+B \right)-\sin \left( A-B \right)=2\cos A\sin B$ from trigonometry.