Question

# Integrate the following integral: $\int{\sin 4x\cos 7xdx}$.

Hint: First, multiply and divide by 2. Take the 2 in the denominator outside of the integral as it is a constant. Next, use the property sin (A + B) = 2sin(A)cos(B) to rewrite it as $\dfrac{1}{2}\int{\left( \sin \left( 11x \right)-\sin \left( 3x \right) \right)dx}$. Now solve this to get the final answer.

In this question, we have to evaluate the following integral: $\int{\sin 4x\cos 7xdx}$.
To solve this, let us first multiply and divide by 2. Doing this, we will get the following:
$\int{\dfrac{2}{2}\sin 4x\cos 7xdx}$
Now, we will take the 2 in the denominator outside of the integral as it is a constant and a constant can be taken outside the integral.
So, we will get the following:
$\dfrac{1}{2}\int{2\sin 4x\cos 7xdx}$ …(1)
Now, we already know that for two angles, say A and B, the sum of the sine of the angle (A + B) and the sine of the angle (A – B) is equal to double of the product of sine of angle A and the cosine of angle B.
i.e. sin (A + B) = 2sin(A)cos(B)
Using this property, we can rewrite (2 sin 4x cos 7x) as:
sin (4x + 7x) + sin (4x – 7x)
sin (11x) + sin (-3x)
Now, we also know that for an angle A. the sine of negative of angle A is equal to the negative of sine of angle A, i.e. sin (-A) = -sin (A).
Using this property, we will get the following:
sin (11x) – sin (3x)
Substituting this in equation (1), we will get the following:
$\dfrac{1}{2}\int{2\sin 4x\cos 7xdx}$
$\dfrac{1}{2}\int{\left( \sin \left( 11x \right)-\sin \left( 3x \right) \right)dx}$
$\dfrac{1}{2}\left[ \dfrac{-\cos 11x}{11}+\dfrac{\cos 3x}{3} \right]+c$
$\dfrac{-\cos 11x}{22}+\dfrac{\cos 3x}{6}+c$
Hence, $\int{\sin 4x\cos 7xdx}=\dfrac{-\cos 11x}{22}+\dfrac{\cos 3x}{6}+c$