Question

Find \[\int {\dfrac{{{{\sin }^2}x - {{\cos }^2}x}}{{\sin x\cos x}}dx} \] .

Answer 1

Hint: You can see it clearly that the question is based on calculus. You may find calculus scary, but it is not. In this problem, we need to find the integration of \[\dfrac{{{{\sin }^2}x - {{\cos }^2}x}}{{\sin x\cos x}}dx\] . You should know the formula of integration of trigonometric ratios and also basic integration formulae before solving this problem. So lets find out the simple answer of the scary question.

Step wise solution:
Given data: To find out the value of \[\int {\dfrac{{{{\sin }^2}x - {{\cos }^2}x}}{{\sin x\cos x}}dx} \] .
To simplify the function \[\dfrac{{{{\sin }^2}x - {{\cos }^2}x}}{{\sin x\cos x}}\]. You must be wondering how to proceed now. Let's make It easy for you:
Suppose, \[I = \int {\dfrac{{{{\sin }^2}x - {{\cos }^2}x}}{{\sin x\cos x}}dx} \]
Let us firstly, expand the function \[\dfrac{{{{\sin }^2}x - {{\cos }^2}x}}{{\sin x\cos x}}\] .
Now, \[\dfrac{{{{\sin }^2}x - {{\cos }^2}x}}{{\sin x\cos x}}\] can be expanded as,

 \[ = \dfrac{{{{\sin }^2}x}}{{\sin x\cos x}} - \dfrac{{{{\cos }^2}x}}{{\sin x\cos x}}\\
 = \dfrac{{\sin x}}{{\cos x}} - \dfrac{{\cos x}}{{\sin x}} \]

Now, given integration can be written as:

\[ I = \int {\dfrac{{{{\sin }^2}x - {{\cos }^2}x}}{{\sin x\cos x}}dx} \\ \]
\[ \Rightarrow I = \int {\dfrac{{\sin x}}{{\cos x}}dx} - \int {\dfrac{{\cos x}}{{\sin x}} dx} \\ \]
 \[ \int {(A + B)dx = \int {Adx} + \int {Bdx} } \\ \]
We will complete the values of \[\int {\dfrac{{\sin x}}{{\cos x}}dx} \,\,and\,\,\int {\dfrac{{\cos x}}{{\sin x}}} dx\] separately;
Firstly, Let us calculate \[\int {\dfrac{{\sin x}}{{\cos x}}dx} \,\]
Suppose, cosx be equal to u.
\[i.e.:cosx = u\]
Differentiating both sides with respect to x, we get
\[
\dfrac{d}{{dx}}cosx = \dfrac{{du}}{{dx}}\\
 \Rightarrow - \sin x[\dfrac{d}{{dx}}(x)] = \dfrac{{du}}{{dx}}\\
i.e.: - \sin x = \dfrac{{du}}{{dx}}\\
 \Rightarrow \sin x\,dx = - du
\]
Now, \[\int {\dfrac{{\sin x}}{{\cos x}}dx} \,\] can be written as
\[
\int {\dfrac{{\sin x}}{{\cos x}}dx} \, = \int {(\dfrac{{ - 1}}{u})dx} \{ u = \cos x,\, - du = \sin x\,dx\\\]
\[ \Rightarrow \int {\dfrac{{\sin x}}{{\cos x}}dx} \, = - \int {\dfrac{1}{u}} \,dx \\
\]
We know that, \[\int {\dfrac{1}{x}} \,dx\] is given as \[\int {\dfrac{1}{x}} \,dx = \ln \left| x \right| + C\]
Hence,
\[
 - \int {\dfrac{1}{u}} \,dx = - [\ln \left| u \right| + {C_1}]\\ \]
 \[ \Rightarrow - \int {\dfrac{1}{u}} \,dx = - \ln \left| u \right| - {C_1}\\ \]
 \[ \Rightarrow \int {\dfrac{{\sin x}}{{\cos x}}dx} \, = - \ln \left| {\cos x} \right| - {C_1} .....(i) \\ \]

Where C1= Constant of integration
Similarly, for \[\int {\dfrac{{\cos x}}{{\sin x}}dx} \,\] ,
\[ \int {\dfrac{{\cos x}}{{\sin x}}dx} \, = \int {\dfrac{1}{v}} \,dv [v = \sin x,dv = \cos x\,dx]\]
Since, \[\int {\dfrac{1}{v}} \,dv\] is equal to \[\ln \left| v \right| + {C_2} \]
Where C2= Constant of integration
\[
 \Rightarrow \int {\dfrac{{\cos x}}{{\sin x}}dx} \, = \ln \left| v \right| + {C_2}\\ \]
\[ \Rightarrow \int {\dfrac{{\cos x}}{{\sin x}}dx} \, = \ln \left| {\sin x} \right| + {C_2} .....(ii)\\
\]
To calculate the required value of I.
Here,
\[
 \Rightarrow I = \int {\dfrac{{\sin x}}{{\cos x}}dx} - \int {\dfrac{{\cos x}}{{\sin x}}} dx\\ \]


Putting the values of equation(i) and (ii), we can write I as,
\[
 \Rightarrow I\, = ( - \ln \left| {\cos x} \right| + {C_1}) - (\ln \left| {\sin x} \right| + {C_2})\\
 \Rightarrow I\, = - \ln \left| {\cos x} \right| + {C_1} - \ln \left| {\sin x} \right| - {C_2}\\
 \Rightarrow I\, = - [\ln \left| {\cos x} \right| + \ln \left| {\sin x} \right| + ({C_2} + {C_1})] \\ \]

Since, C1 and C2 are two constants of integration, -(C1+C2) is also constant. Let –(C1+C2) be.
\[ \Rightarrow I\, = - (\ln \left| {\cos x} \right| + \ln \left| {\sin x} \right|) + C\]
According to the properties of logarithm,
\[\ln \left| A \right| + \ln \left| B \right|\] is equal to \[ln\left| {A \cdot B} \right|\]
Applying the properties in I, we get
\[
 \Rightarrow I\, = - \ln \left| {\cos x + \sin x} \right| + C\\\]
i.e.: \[\int {\dfrac{{{{\sin }^2}x - {{\cos }^2}x}}{{\sin x\cos x}}dx} = - \ln \left| {\cos x + \sin x} \right| + C \\ \]

Required value of the given integration.


Note: In this problem, there are trigonometric ratios, logarithm and also most importantly integration. You should grasp knowledge about all of the above mentioned topics to solve sums like this. Also , students, you can try taking and solve and not expanding the function. It will help you in solving questions without getting lengthy.