Question

Evaluate $\int {\dfrac{{3\sin x + 2\cos x}}{{3\cos x + 2\sin x}}dx}$

Answer 1

Hint : The given question requires us to integrate a rational trigonometric function in variable x with respect to x. Now, to integrate the function given to us, we transform the numerator of the function into two parts, first being equal to the denominator and second one being the derivative of the denominator. So, we convert the given integral into two different integrals that are relatively easier to compute. We must know integration by substitution in order to solve such types of questions.

Complete step by step solution:
The given question requires us to integrate a rational trigonometric function $\dfrac{{3\sin x + 2\cos x}}{{3\cos x + 2\sin x}}$ in variable x.
We can convert the given integral $\int {\dfrac{{3\sin x + 2\cos x}}{{3\cos x + 2\sin x}}dx}$ into two different integrals by splitting up the function into two parts so as to make the integration of the function easier.
So, we split up the numerator of the function $\dfrac{{3\sin x + 2\cos x}}{{3\cos x + 2\sin x}}$ into two parts. The first part is the denominator and second part is the derivative of the denominator.
Now, we know that the derivative of $\sin x$ is $\cos x$ and derivative of $\cos x$ is $- \sin x$ . So, we get the derivative of the denominator $\left( {3\cos x + 2\sin x} \right)$ as $\left( { - 3\sin x + 2\cos x} \right)$ . So, we get,
$3\sin x + 2\cos x = A\left( {3\cos x + 2\sin x} \right) + B\left( { - 3\sin x + 2\cos x} \right)$
Now, we have to find the values of A and B in the above equation. So, separating the sine and cosine terms I the right side of the equation, we get,
$\Rightarrow 3\sin x + 2\cos x = \cos x\left( {3A + 2B} \right) + \sin x\left( {2A - 3B} \right)$
Now, comparing the coefficients of sine and cosine on both sides of the equation, we get,
$3A + 2B = 2 - - - - - \left( 1 \right)$
$2A - 3B = 3 - - - - - \left( 2 \right)$
Now, we have to solve these two equations so as to find the values of A and B.
Multiplying both sides of the equation $\left( 1 \right)$ by $3$ , we get,
$\Rightarrow 9A + 6B = 6 - - - - - \left( 3 \right)$
Multiplying both sides of the equation $\left( 2 \right)$ by $2$ , we get,
$\Rightarrow 4A - 6B = 6 - - - - - \left( 4 \right)$
Now, adding the equations $\left( 3 \right)$ and $\left( 4 \right)$ , we get,
$\Rightarrow 9A + 6B + 4A - 6B = 6 + 6$
Cancelling the like terms with opposite signs, we get,
$\Rightarrow 9A + 4A = 12$
$\Rightarrow 13A = 12$
Dividing both sides of the equation by $13$ , we get,
$\Rightarrow 13A = 12$
$\Rightarrow A = \dfrac{{12}}{{13}}$
So, we get the value of A as $\left( {\dfrac{{12}}{{13}}} \right)$ . Now, substituting the value of A in the equation $\left( 1 \right)$ to find the value of B, we get,
$\Rightarrow 3\left( {\dfrac{{12}}{{13}}} \right) + 2B = 2$
Opening the bracket, we get,
$\Rightarrow \dfrac{{36}}{{13}} + 2B = 2$
Shifting all the constants to the right side of the equation, we get,
$\Rightarrow 2B = 2 - \dfrac{{36}}{{13}}$
$\Rightarrow 2B = \dfrac{{26 - 36}}{{13}} = \dfrac{{ - 10}}{{13}}$
Dividing both sides of the equation by $2$ , we get,
$\Rightarrow B = \dfrac{{ - 5}}{{13}}$
Hence, we get, $3\sin x + 2\cos x = \dfrac{{12}}{{13}}\left( {3\cos x + 2\sin x} \right) - \dfrac{5}{{13}}\left( { - 3\sin x + 2\cos x} \right)$
So, we have split the numerator into two parts. Now, substituting this in the integral, we get,
$\Rightarrow \int {\dfrac{{3\sin x + 2\cos x}}{{3\cos x + 2\sin x}}dx} = \int {\dfrac{{\dfrac{{12}}{{13}}\left( {3\cos x + 2\sin x} \right) - \dfrac{5}{{13}}\left( { - 3\sin x + 2\cos x} \right)}}{{3\cos x + 2\sin x}}dx}$
Splitting the integral into two parts, we get,
$\Rightarrow \int {\dfrac{{12}}{{13}}\dfrac{{\left( {3\cos x + 2\sin x} \right)}}{{\left( {3\cos x + 2\sin x} \right)}}dx} - \int {\dfrac{5}{{13}}\dfrac{{\left( { - 3\sin x + 2\cos x} \right)}}{{\left( {3\cos x + 2\sin x} \right)}}dx}$
Taking constants out of the integral and cancelling the common factors in numerator and denominator, we get,
$\Rightarrow \dfrac{{12}}{{13}}\int {1dx} - \dfrac{5}{{13}}\int {\dfrac{{\left( { - 3\sin x + 2\cos x} \right)}}{{\left( {3\cos x + 2\sin x} \right)}}dx}$
Now, we know that integration of $1$ with respect to x is x. So, we get,
$\Rightarrow \dfrac{{12}}{{13}}x - \dfrac{5}{{13}}\int {\dfrac{{\left( { - 3\sin x + 2\cos x} \right)}}{{\left( {3\cos x + 2\sin x} \right)}}dx}$
Now, to solve the second integral, we substitute the denominator as t. So, $\left( {3\cos x + 2\sin x} \right) = t$ .
Differentiating both sides of the equation, we get,
$dt = \left( { - 3\sin x + 2\cos x} \right)dx$
So, we get the second integral in terms of t as,
$\Rightarrow \dfrac{{12}}{{13}}x - \dfrac{5}{{13}}\int {\dfrac{{dt}}{t}}$
Now, we know that integration of $\dfrac{1}{t}$ with respect to $dt$ is $\log \left| t \right|$ .
$\Rightarrow \dfrac{{12}}{{13}}x - \dfrac{5}{{13}}\log \left| t \right| + c$ , where c is any arbitrary constant.
Substituting back the value of t in the integral, we get,
$\Rightarrow \dfrac{{12}}{{13}}x - \dfrac{5}{{13}}\log \left| {3\cos x + 2\sin x} \right| + c$
So, we get the value of integral $\int {\dfrac{{3\sin x + 2\cos x}}{{3\cos x + 2\sin x}}dx}$ as $\dfrac{{12}}{{13}}x - \dfrac{5}{{13}}\log \left| {3\cos x + 2\sin x} \right| + c$ , where c is an arbitrary constant.
So, option (B) is correct.
So, the correct answer is “Option B”.

Note : The indefinite integrals of certain functions may have more than one answer in different forms. We must remember some special types of integration methods like the one used in the question to solve complex problems with ease. However, all these forms are correct and interchangeable into one another. We should not forget to add an arbitrary constant to the answer of the indefinite integral. Indefinite integral gives us the family of curves as we don’t know the exact value of the arbitrary constant.