Question

Evaluate the value of the integral $\int {\dfrac{{\sin x + 4\sin 3x + 6\sin 5x + 3\sin 7x}}{{\sin 2x + 3\sin 4x + 3\sin 6x}}} dx$.

Answer 1

Hint: Here, we will proceed by splitting the terms in the numerator of the given integral in such a way that the formulas $\sin {\text{A}} + \sin {\text{B}} = 2\sin \left( {\dfrac{{{\text{A}} + {\text{B}}}}{2}} \right)\cos \left( {\dfrac{{{\text{A}} - {\text{B}}}}{2}} \right)$ and $\cos {\text{A}} + \cos {\text{B}} = 2\cos \left( {\dfrac{{{\text{A}} + {\text{B}}}}{2}} \right)\cos \left( {\dfrac{{{\text{A}} - {\text{B}}}}{2}} \right)$ can be used to convert the numerator in terms of the denominator of the given integral.

Complete step-by-step answer:
Let us suppose the integral ${\text{I}} = \int {\left( {\dfrac{{\sin x + 4\sin 3x + 6\sin 5x + 3\sin 7x}}{{\sin 2x + 3\sin 4x + 3\sin 6x}}} \right)} dx{\text{ }} \to {\text{(1)}}$
Let us consider the numerator of the integral mentioned in equation (1), we get
$\sin x + 4\sin 3x + 6\sin 5x + 3\sin 7x = \sin x + \left( {4\sin 3x + 4\sin 5x} \right) + \left( {2\sin 5x + 2\sin 7x} \right) + \sin 7x \\\ \Rightarrow \sin x + 4\sin 3x + 6\sin 5x + 3\sin 7x = \left( {\sin x + \sin 7x} \right) + 4\left( {\sin 3x + \sin 5x} \right) + 2\left( {\sin 5x + \sin 7x} \right){\text{ }} \to {\text{(2)}} \\\$
As we know that $\sin {\text{A}} + \sin {\text{B}} = 2\sin \left( {\dfrac{{{\text{A}} + {\text{B}}}}{2}} \right)\cos \left( {\dfrac{{{\text{A}} - {\text{B}}}}{2}} \right)$
Using the above formula, equation (2) becomes

$$\Rightarrow \sin x + 4\sin 3x + 6\sin 5x + 3\sin 7x = 2\sin \left( {\dfrac{{x + 7x}}{2}} \right)\cos \left( {\dfrac{{x - 7x}}{2}} \right) + 4\left[ {2\sin \left( {\dfrac{{3x + 5x}}{2}} \right)\cos \left( {\dfrac{{3x - 5x}}{2}} \right)} \right] + 2\left[ {2\sin \left( {\dfrac{{5x + 7x}}{2}} \right)\cos \left( {\dfrac{{5x - 7x}}{2}} \right)} \right] \\\ \Rightarrow \sin x + 4\sin 3x + 6\sin 5x + 3\sin 7x = 2\sin \left( {\dfrac{{8x}}{2}} \right)\cos \left( {\dfrac{{ - 6x}}{2}} \right) + 4\left[ {2\sin \left( {\dfrac{{8x}}{2}} \right)\cos \left( {\dfrac{{ - 2x}}{2}} \right)} \right] + 2\left[ {2\sin \left( {\dfrac{{12x}}{2}} \right)\cos \left( {\dfrac{{ - 2x}}{2}} \right)} \right] \\\ \Rightarrow \sin x + 4\sin 3x + 6\sin 5x + 3\sin 7x = 2\sin \left( {4x} \right)\cos \left( { - 3x} \right) + 8\sin \left( {4x} \right)\cos \left( { - x} \right) + 4\sin \left( {6x} \right)\cos \left( { - x} \right){\text{ }} \to (3) \\\$$

Using the formula $\cos \left( { - \theta } \right) = \cos \theta$ in equation (3), we get

$$\Rightarrow \sin x + 4\sin 3x + 6\sin 5x + 3\sin 7x = 2\sin 4x\cos 3x + 8\sin 4x\cos x + 4\sin 6x\cos x \\\ \Rightarrow \sin x + 4\sin 3x + 6\sin 5x + 3\sin 7x = \left( {2\sin 4x\cos 3x + 2\sin 4x\cos x} \right) + 2\sin 4x\cos x + \left( {4\sin 4x\cos x + 4\sin 6x\cos x} \right) \\\$$

By taking 2sin4x and 4cosx common from the first two and the last two terms in the RHS of the above equation, we get
$\Rightarrow \sin x + 4\sin 3x + 6\sin 5x + 3\sin 7x = 2\sin 4x\left( {\cos 3x + \cos x} \right) + 2\sin 4x\cos x + 4\cos x\left( {\sin 4x + \sin 6x} \right){\text{ }} \to {\text{(4)}}$
Using the formulas $\sin {\text{A}} + \sin {\text{B}} = 2\sin \left( {\dfrac{{{\text{A}} + {\text{B}}}}{2}} \right)\cos \left( {\dfrac{{{\text{A}} - {\text{B}}}}{2}} \right)$ and $\cos {\text{A}} + \cos {\text{B}} = 2\cos \left( {\dfrac{{{\text{A}} + {\text{B}}}}{2}} \right)\cos \left( {\dfrac{{{\text{A}} - {\text{B}}}}{2}} \right)$ in the RHS of equation (4), we get

$$\Rightarrow \sin x + 4\sin 3x + 6\sin 5x + 3\sin 7x = 2\sin 4x\left[ {2\cos \left( {\dfrac{{3x + x}}{2}} \right)\cos \left( {\dfrac{{3x - x}}{2}} \right)} \right] + 2\sin 4x\cos x + 4\cos x\left[ {2\sin \left( {\dfrac{{4x + 6x}}{2}} \right)\cos \left( {\dfrac{{4x - 6x}}{2}} \right)} \right] \\\ \Rightarrow \sin x + 4\sin 3x + 6\sin 5x + 3\sin 7x = 2\sin 4x\left[ {2\cos \left( {\dfrac{{4x}}{2}} \right)\cos \left( {\dfrac{{2x}}{2}} \right)} \right] + 2\sin 4x\cos x + 4\cos x\left[ {2\sin \left( {\dfrac{{10x}}{2}} \right)\cos \left( {\dfrac{{ - 2x}}{2}} \right)} \right] \\\ \Rightarrow \sin x + 4\sin 3x + 6\sin 5x + 3\sin 7x = 2\sin 4x\left[ {2\cos \left( {2x} \right)\cos \left( x \right)} \right] + 2\sin 4x\cos x + 4\cos x\left[ {2\sin \left( {5x} \right)\cos \left( { - x} \right)} \right]{\text{ }} \to {\text{(5)}} \\\$$

Using the formula $\cos \left( { - \theta } \right) = \cos \theta$ in equation (5), we get

$$\Rightarrow \sin x + 4\sin 3x + 6\sin 5x + 3\sin 7x = 4\sin 4x\cos 2x\cos x + 2\sin 4x\cos x + 8\cos x\sin 5x\cos x \\\ \Rightarrow \sin x + 4\sin 3x + 6\sin 5x + 3\sin 7x = 4\sin 4x\cos 2x\cos x + 2\sin 4x\cos x + 8\cos x\sin 5x\cos x \\\$$

Now taking 2cox common from all the terms in the RHS of the above equation, we get

$$\Rightarrow \sin x + 4\sin 3x + 6\sin 5x + 3\sin 7x = 2\cos x\left( {2\sin 4x\cos 2x + \sin 4x + 4\sin 5x\cos x} \right) \\\ \Rightarrow \sin x + 4\sin 3x + 6\sin 5x + 3\sin 7x = 2\cos x\left( {\left[ {2\sin \left( {\dfrac{{6x + 2x}}{2}} \right)\cos \left( {\dfrac{{6x - 2x}}{2}} \right)} \right] + \sin 4x + 2\left[ {2\sin \left( {\dfrac{{6x + 4x}}{2}} \right)\cos \left( {\dfrac{{6x - 4x}}{2}} \right)} \right]} \right) \\\$$

Using the formula $2\sin \left( {\dfrac{{{\text{A}} + {\text{B}}}}{2}} \right)\cos \left( {\dfrac{{{\text{A}} - {\text{B}}}}{2}} \right) = \sin {\text{A}} + \sin {\text{B}}$, the above equation becomes

$$\Rightarrow \sin x + 4\sin 3x + 6\sin 5x + 3\sin 7x = 2\cos x\left( {\left[ {\sin 6x + \sin 2x} \right] + \sin 4x + 2\left[ {\sin 6x + \sin 4x} \right]} \right) \\\ \Rightarrow \sin x + 4\sin 3x + 6\sin 5x + 3\sin 7x = 2\cos x\left( {\sin 6x + \sin 2x + \sin 4x + 2\sin 6x + 2\sin 4x} \right) \\\ \Rightarrow \sin x + 4\sin 3x + 6\sin 5x + 3\sin 7x = 2\cos x\left( {\sin 2x + 3\sin 4x + 3\sin 6x} \right){\text{ }} \to {\text{(6)}} \\\$$

Now, substitute the equation (6) in equation (1), we get
$\Rightarrow {\text{I}} = \int {\left( {\dfrac{{2\cos x\left( {\sin 2x + 3\sin 4x + 3\sin 6x} \right)}}{{\sin 2x + 3\sin 4x + 3\sin 6x}}} \right)} dx \\\ \Rightarrow {\text{I}} = \int {\left( {2\cos x} \right)} dx \\\ \Rightarrow {\text{I}} = 2\int {\cos x} dx \\\ \Rightarrow {\text{I}} = 2\left[ {\sin x} \right] + c \\\ \Rightarrow {\text{I}} = 2\sin x + c \\\$
Where c is the constant of integration.
Therefore, the value of the integral $\int {\dfrac{{\sin x + 4\sin 3x + 6\sin 5x + 3\sin 7x}}{{\sin 2x + 3\sin 4x + 3\sin 6x}}} dx$ is $\left( {2\sin x + c} \right)$.

Note: In this particular problem, we have written the terms in the numerator of the given integral in such a way that the multiples of the sine trigonometric function becomes same and then by taking that multiple common, we can use the formula $\sin {\text{A}} + \sin {\text{B}} = 2\sin \left( {\dfrac{{{\text{A}} + {\text{B}}}}{2}} \right)\cos \left( {\dfrac{{{\text{A}} - {\text{B}}}}{2}} \right)$ and when all the terms are reduced we are using the inverse of this formula so that some the given integral is simplified into an easy integral.