Question

Integration of : \[\int {\dfrac{{\cos x - \cos 2x}}{{1 - \cos x}}} dx\]?

Answer 1

Hint: To solve this integration, we will first convert the whole expression in terms of \[{\cos ^n}x\]. For that, we will first convert \[\cos 2x\] and write it in terms of \[\cos x\] using the formula \[\cos 2x = 2{\cos ^2}x - 1\]. After that, we will have a quadratic equation in the numerator. Now, we will simplify the numerator by factorising and then cancel out some of the terms from numerator and denominator. After cancellation, we will be left with some terms in the numerator and then we will integrate them using the integration formulas and obtain our answer.

Complete step by step answer:
We need to integrate \[\int {\dfrac{{\cos x - \cos 2x}}{{1 - \cos x}}} dx\]
Let \[I = \int {\dfrac{{\cos x - \cos 2x}}{{1 - \cos x}}} dx\]
Now, convert this expression in terms of \[{\cos ^n}x\].
Using the formula \[\cos 2x = 2{\cos ^2}x - 1\], we have
\[ \Rightarrow I = \int {\dfrac{{\cos x - \left( {2{{\cos }^2}x - 1} \right)}}{{1 - \cos x}}} dx\]
Opening the brackets, we get
\[ \Rightarrow I = \int {\dfrac{{\cos x - 2{{\cos }^2}x + 1}}{{1 - \cos x}}} dx\]

Taking the negative sign common from the numerator and denominator respectively, we get
\[ \Rightarrow I = \int {\dfrac{{\left( - \right)\left( { - \cos x + 2{{\cos }^2}x - 1} \right)}}{{\left( - \right)\left( { - 1 + \cos x} \right)}}} dx\]
Cancelling out the negative sign, we get
\[ \Rightarrow I = \int {\dfrac{{\left( { - \cos x + 2{{\cos }^2}x - 1} \right)}}{{\left( { - 1 + \cos x} \right)}}} dx\]
Rearranging the terms, we get
\[ \Rightarrow I = \int {\dfrac{{\left( {2{{\cos }^2}x - \cos x - 1} \right)}}{{\left( {\cos x - 1} \right)}}} dx\]
We will factorise the numerator using Splitting the middle term formula.
We can write \[\cos x\] as \[2\cos x - \cos x\]
Hence, writing \[\cos x\] as \[2\cos x - \cos x\], we get
\[ \Rightarrow I = \int {\dfrac{{\left( {2{{\cos }^2}x - \left( {2\cos x - \cos x} \right) - 1} \right)}}{{\left( {\cos x - 1} \right)}}} dx\]

Now, opening the brackets in the numerator, we get
\[ \Rightarrow I = \int {\dfrac{{\left( {2{{\cos }^2}x - 2\cos x + \cos x - 1} \right)}}{{\left( {\cos x - 1} \right)}}} dx\]
Now, combing first two terms and the last two terms together in the numerator, we get
\[ \Rightarrow I = \int {\dfrac{{\left( {2{{\cos }^2}x - 2\cos x} \right) + \left( {\cos x - 1} \right)}}{{\left( {\cos x - 1} \right)}}} dx\]
Taking \[2\cos x\] common from the first two terms, we get
\[ \Rightarrow I = \int {\dfrac{{\left( {2\cos x} \right)\left( {\cos x - 1} \right) + \left( {\cos x - 1} \right)}}{{\left( {\cos x - 1} \right)}}} dx\]
Using \[a = \left( 1 \right)\left( a \right)\], we can write \[\cos x - 1\] as \[\left( 1 \right)\left( {\cos x - 1} \right)\]. Hence, our expression becomes
\[ \Rightarrow I = \int {\dfrac{{\left( {2\cos x} \right)\left( {\cos x - 1} \right) + \left( 1 \right)\left( {\cos x - 1} \right)}}{{\left( {\cos x - 1} \right)}}} dx\]

Now, taking \[\left( {\cos x - 1} \right)\] common from the numerator, we get
\[ \Rightarrow I = \int {\dfrac{{\left( {\cos x - 1} \right)\left( {2\cos x + 1} \right)}}{{\left( {\cos x - 1} \right)}}} dx\]
Now, cancelling the term \[\left( {\cos x - 1} \right)\] from the numerator and denominator, we get
\[ \Rightarrow I = \int {\left( {2\cos x + 1} \right)} dx\]
Now, using \[\int {\left( {f\left( x \right) + g\left( x \right)} \right)} dx = \int {f\left( x \right)} dx + \int {g\left( x \right)} dx\], we get
\[ \Rightarrow I = \int {\left( {2\cos x} \right)} dx + \int {\left( 1 \right)} dx\]
Using the property \[\int {\left( {c.f\left( x \right)} \right)} dx = c\int {\left( {f\left( x \right)} \right)} dx\], where \[c\] is a constant, we get
\[ \Rightarrow I = 2\int {\left( {\cos x} \right)} dx + \int {\left( 1 \right)} dx\]

Now, using \[\int {\left( {\cos x} \right)} dx = \sin x + c\] and \[\int {\left( m \right)} dx = mx + c\], where \[m\] and \[c\] are the constant terms, we get
\[ \Rightarrow I = 2\left( {\sin x + {c_1}} \right) + \left( {1.x} \right) + {c_2}\], where \[{c_1}\] and \[{c_2}\] are constant terms.
Opening the brackets, we get
\[ \Rightarrow I = 2\sin x + 2{c_1} + x + {c_2}\], where \[{c_1}\] and \[{c_2}\] are constant terms.
Now combining the constant terms, we get
\[ \Rightarrow I = 2\sin x + x + \left( {2{c_1} + {c_2}} \right)\], where \[{c_1}\] and \[{c_2}\] are constant terms.
\[ \Rightarrow I = 2\sin x + x + c\], where \[c = \left( {2{c_1} + {c_2}} \right)\], where \[{c_1}\] and \[{c_2}\] are constant terms.
Hence, we get \[I = 2\sin x + x + c\], where \[c\] is a constant term.

Therefore, \[\int {\dfrac{{\cos x - \cos 2x}}{{1 - \cos x}}} dx = 2\sin x + x + c\], where \[c\] is a constant term.

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.