Question

Find the value of $\int {\dfrac{1}{{{{\left( {2\sin x + 3\cos x} \right)}^2}}}dx}$.
A) $\dfrac{1}{{2\left( {2\tan x - 3} \right)}}$
B) $\dfrac{1}{{2\left( {\tan x + 3} \right)}}$
C) $- \dfrac{1}{{2\left( {2\tan x + 3} \right)}}$
D) $- \dfrac{1}{{2\left( {2\tan x - 3} \right)}}$

Answer 1

Here, we will first divide the numerator and denominator by ${\cos ^2}\left( x \right)$ in the above equation and use the value $\dfrac{1}{{\cos x}} = \sec x$ and $\dfrac{{\sin x}}{{\cos x}} = \tan x$ in the obtained expression. Then we will take $u = \tan x$, then differentiating it with respect to $x$ and substitute $u = \tan x$ and take $v = 2u + 3$, then differentiating it with respect to $u$. Then we will substitute $v = 2u + 3$and apply the power rule, $\int {{v^n}dv} = \dfrac{{{v^{n + 1}}}}{{n + 1}}$ in the equation and then substitute back the values to find the required value.

Complete step by step solution:
We are given that
$\int {\dfrac{1}{{{{\left( {2\sin x + 3\cos x} \right)}^2}}}dx}$
Dividing the numerator and denominator by ${\cos ^2}\left( x \right)$ in the above equation, we get

$$\Rightarrow \int {\dfrac{{\dfrac{1}{{{{\cos }^2}x}}}}{{\dfrac{{{{\left( {2\sin x + 3\cos x} \right)}^2}}}{{{{\cos }^2}x}}}}dx} \\\ \Rightarrow \int {\dfrac{{\dfrac{1}{{{{\cos }^2}x}}}}{{{{\left( {\dfrac{{2\sin x}}{{\cos x}} + \dfrac{{3\cos x}}{{\cos x}}} \right)}^2}}}dx} \\\ \Rightarrow \int {\dfrac{{\dfrac{1}{{{{\cos }^2}x}}}}{{{{\left( {\dfrac{{2\sin x}}{{\cos x}} + 3} \right)}^2}}}dx} \\\$$

Using the value $\dfrac{1}{{\cos x}} = \sec x$ and $\dfrac{{\sin x}}{{\cos x}} = \tan x$ in the above equation, we get
$\Rightarrow \int {\dfrac{{{{\sec }^2}x}}{{{{\left( {2\tan x + 3} \right)}^2}}}dx} {\text{ ......eq.(1)}}$
Taking $u = \tan x$, then differentiating it with respect to $x$, we get
$\Rightarrow \dfrac{{du}}{{dx}} = {\sec ^2}x$
Cross-multiplying the above equation, we get
$\Rightarrow du = {\sec ^2}xdx$
Substituting $u = \tan x$and the above equation in the equation (1), we get
$\Rightarrow \int {\dfrac{1}{{{{\left( {2u + 3} \right)}^2}}}du} {\text{ ......eq.(2)}}$
Taking $v = 2u + 3$, then differentiating it with respect to $u$, we get
$\Rightarrow \dfrac{{dv}}{{du}} = 2$
Cross-multiplying the above equation, we get
$\Rightarrow du = \dfrac{1}{2}dv$
Substituting $v = 2u + 3$and the above equation in the equation (2), we get
$\Rightarrow \dfrac{1}{2}\int {\dfrac{1}{{{v^2}}}du}$
Applying the power rule, $\int {{v^n}dv} = \dfrac{{{v^{n + 1}}}}{{n + 1}}$ in the above equation, we get

$$\Rightarrow \left( {\dfrac{{{v^{ - 2 + 1}}}}{{\left( { - 2 + 1} \right)}}} \right) \\\ \Rightarrow - {v^{ - 1}} \\\ \Rightarrow - \dfrac{1}{v} \\\$$

Substituting $2u + 3 = v$and the above equation, we get
$\Rightarrow - \dfrac{1}{{2u + 3}}$
Substituting $\tan x = u$in the above equation, we get
$\Rightarrow - \dfrac{1}{{2\left( {2\tan x + 3} \right)}}$

Hence, option C is correct.

Note:
We need to know that while finding the value of indefinite integral, we have to add the constant in the final answer or else the answer will be incomplete. But since the options did not have any constant so we do not have to write it here. We have to be really thorough with the integrations and differentiation of the functions. The key point in this question is to use the substitution and integration rule,$\int {{v^n}dv} = \dfrac{{{v^{n + 1}}}}{{n + 1}}$ to solve this problem. Do not forget that many integrals can be evaluated in multiple ways and so more than one technique may be used on it, but this problem can only be solved by parts.