Question

Evaluate \[{\int {\left( {\dfrac{x}{{x\sin x + \cos x}}} \right)} ^2}dx\].
A. \[\tan x - \dfrac{{x\sec x}}{{x\sin x + \cos x}} + C\]
B. \[\sec x - \dfrac{{x\tan x}}{{x\sin x + \cos x}} + C\]
C. \[\sec x + \dfrac{{x\tan x}}{{x\sin x + \cos x}} + C\]
D. \[\tan x + \dfrac{{x\sec x}}{{x\sin x + \cos x}} + C\]

Answer 1

Hint: First we will simplify the numerator of the integrating term. Rewrite the numerator of the integrating term as \[x\cos x \cdot x\sec x\]. We will apply the \[uv\] formula of integration and then we will assume \[t = x\sin x + \cos x\] and find the derivative of it to find \[{\int {\left( {\dfrac{x}{{x\sin x + \cos x}}} \right)} ^2}dx\].

Formula used
\[\cos x \cdot \sec x = 1\]
\[\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}\]
\[\dfrac{d}{{dx}}\sin x = \cos x\]
\[\dfrac{d}{{dx}}\cos x = - \sin x\]
\[\int {\dfrac{1}{{{t^2}}}} dt = - \dfrac{1}{t} + C\]
\[\int {uvdx} = u\int {vdx} - \int {\left[ {\dfrac{d}{{dx}}u\int {vdx} } \right]} dx\]
\[\int {{{\sec }^2}xdx} = \tan x + C\]

Complete step by step solution:
Given integration is \[{\int {\left( {\dfrac{x}{{x\sin x + \cos x}}} \right)} ^2}dx\].
Now rewrite the numerator
\[ = \int {\dfrac{{{x^2}}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}} dx\]
Rewrite the numerator of the integrating term as \[x\cos x \cdot x\sec x\]
\[ = \int {\dfrac{{x\cos x \cdot x\sec x}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}} dx\]
\[ = \int {x\sec x \cdot \dfrac{{x\cos x}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}} dx\]
Apply \[\int {uvdx} = u\int {vdx} - \int {\left[ {\dfrac{d}{{dx}}u\int {vdx} } \right]} dx\]
\[ = x\sec x \cdot \int {\dfrac{{x\cos x}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}} dx - \int {\left[ {\dfrac{d}{{dx}}\left( {x\sec x} \right)\int {\dfrac{{x\cos x}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}} dx} \right]dx} \] ………(1)
Now calculating \[\int {\dfrac{{x\cos x}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}} dx\]
Let \[x\sin x + \cos x = t\]
Differentiate both sides
\[\left( {x\cos x + \sin x - \sin x} \right)dx = dt\]
\[ \Rightarrow x\cos xdx = dt\]
Now substitute \[x\cos xdx = dt\] and \[x\sin x + \cos x = t\] in \[\int {\dfrac{{x\cos x}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}} dx\]
\[\int {\dfrac{{x\cos x}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}} dx\]
\[ = \int {\dfrac{1}{{{t^2}}}} dt\]
Apply the formula \[\int {\dfrac{1}{{{x^2}}}} dx = - \dfrac{1}{x} + C\]
\[ = - \dfrac{1}{t} + {C_1}\]
Now substituting \[x\sin x + \cos x = t\]
\[ = - \dfrac{1}{{x\sin x + \cos x}} + {C_1}\]
Now calculating the differential \[\dfrac{d}{{dx}}\left( {x\sec x} \right)\]
Applying the formula \[\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}\]
\[\dfrac{d}{{dx}}\left( {x\sec x} \right) = x\dfrac{d}{{dx}}\sec x + \sec x\dfrac{d}{{dx}}x\]
Applying the formula \[\dfrac{d}{{dx}}\sec x = \sec x\tan x\] and \[\dfrac{d}{{dx}}x = 1\]
\[ \Rightarrow \dfrac{d}{{dx}}\left( {x\sec x} \right) = x\sec x\tan x + \sec x \cdot 1\]
\[ \Rightarrow \dfrac{d}{{dx}}\left( {x\sec x} \right) = x\sec x\tan x + \sec x\]
\[ \Rightarrow \dfrac{d}{{dx}}\left( {x\sec x} \right) = x \cdot \dfrac{1}{{\cos x}} \cdot \dfrac{{\sin x}}{{\cos x}} + \dfrac{1}{{\cos x}}\]
\[ \Rightarrow \dfrac{d}{{dx}}\left( {x\sec x} \right) = x\dfrac{{\sin x}}{{{{\cos }^2}x}} + \dfrac{1}{{\cos x}}\]
\[ \Rightarrow \dfrac{d}{{dx}}\left( {x\sec x} \right) = \dfrac{{x\sin x + \cos x}}{{{{\cos }^2}x}}\]
Now putting \[\int {\dfrac{{x\cos x}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}} dx = - \dfrac{1}{{x\sin x + \cos x}} + {C_1}\] and \[\dfrac{d}{{dx}}\left( {x\sec x} \right) = \dfrac{{x\sin x + \cos x}}{{{{\cos }^2}x}}\] in equation (1)
\[{\int {\left( {\dfrac{x}{{x\sin x + \cos x}}} \right)} ^2}dx = x\sec x \cdot \int {\dfrac{{x\cos x}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}} dx - \int {\left[ {\dfrac{d}{{dx}}\left( {x\sec x} \right)\int {\dfrac{{x\cos x}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}} dx} \right]dx} \]
\[ = x\sec x\left( { - \dfrac{1}{{x\sin x + \cos x}}} \right) + \int {\dfrac{{x\sin x + \cos x}}{{{{\cos }^2}x}}\left( { - \dfrac{1}{{x\sin x + \cos x}}} \right)dx} \]
\[ = x\sec x\left( { - \dfrac{1}{{x\sin x + \cos x}}} \right) + \int {{{\sec }^2}xdx} \]
Now applying the formula \[\int {{{\sec }^2}xdx} = \tan x + C\]
\[ = - \dfrac{{x\sec x}}{{x\sin x + \cos x}} + \tan x + C\]
Hence option A is the correct option.

Note: Many students try to solve the integration directly and they do not substitute \[x\cos x \cdot x\sec x\] in numerator. For this reason, they are unable to reach the answer.