Question

Find the value of $\int{\dfrac{3+2\cos x}{2+3\cos x}dx}$:
A. $\left[ \dfrac{\sin x}{2+3\cos x} \right]+C$
B. $\left[ \dfrac{2\cos x}{2+3\sin x} \right]+C$
C. $\left[ \dfrac{2\cos x}{2+3\cos x} \right]+C$
D. $\left[ \dfrac{2\sin x}{2+3\sin x} \right]+C$

Answer 1

Hint: For solving this question you should know about integration of the given trigonometric functions. In this problem we will substitute the method by substituting $\dfrac{\sin x}{3\cos x+2}=t$ and then we will differentiate it. And we put the values in the integral as a form of t and dt and then by solving this we will get a new integral in complete terms of t. Then we will replace t with the substitution term and therefore, we will get our answer.

Complete step by step answer:
According to our question it is asked to us to find the integration of the integral $\dfrac{3+2\cos x}{{{\left( 2+3\cos x \right)}^{2}}}$.
Let $I=\int{\dfrac{3+2\cos x}{{{\left( 2+3\cos x \right)}^{2}}}.dx}$
Let us consider $\dfrac{\sin x}{3\cos x+2}=t$
Now, we will use division rule for differentiation of $\dfrac{\sin x}{3\cos x+2}=t$
Therefore, on differentiation, we will get
$\dfrac{\left( 3\cos x+2 \right)\cos x-\sin x\left( -3\sin x \right)}{{{\left( 3\cos x+2 \right)}^{2}}}dx=dt$
And we will further simplify it. Therefore, we get
$\begin{align}
  & \dfrac{3{{\cos }^{2}}x+2\cos x+3{{\sin }^{2}}x}{{{\left( 3\cos x+2 \right)}^{2}}}dx=dt \\
 & \Rightarrow \dfrac{3+2\cos x}{{{\left( 3\cos x+2 \right)}^{2}}}dx=dt \\
\end{align}$
Now, we will write I in terms of t and dt. Therefore, we get
$I=\int{dt}$
And we will use the basic integration rule. Therefore, we will get
$I=t+c$
And on putting value of t back, we will get
$I=\dfrac{\sin x}{3\cos x+2}+c$


So, the correct answer is “Option A”.

Note: Usually the method of integration by substitution for a function whose derivative is also present in the integral. Doing so, the function simplifies and then the basic formulas of integration can be used to integrate the function.