Question

The \[\int {(1 - \cos x)} \cos e{c^2}x\] \[dx\] equals
A. \[\tan \dfrac{x}{2} + c\]
B. \[ - \cot \dfrac{x}{2} + c\]
C. \[2\tan \dfrac{x}{2} + c\]
D. \[ - 2\cot \dfrac{x}{2} + c\]

Answer 1

Hint: For this type of integration, we need to first open the brackets and then solve the Integration individually i.e. We know,
\[\int {(x - y)dx = \int {xdx - \int {ydx} } } \]
We will then write \[\csc x\] in terms of \[\sin x\]and then solve \[\int {\cos x \times \cos e{c^2}x} \] \[dx\]in terms of \[\sin x\] and \[\cos x\] by substituting \[\sin x\] by \[t\] and then replacing \[\cos x\]\[dx\] by \[dt\].We will then subtract the two integrals in the given order. We need to keep in mind all the formulas for integration and trigonometry. After finding the value of the integral, we need to transform the obtained answer according to the given options.

Complete step by step answer:
So, we have the integral \[\int {(1 - \cos x)} \cos e{c^2}x\] \[dx\]
Opening the brackets, we get,
\[\int {(1 - \cos x)\cos e{c^2}x} \] \[dx\] \[ = \int {(\cos e{c^2}x - (\cos x \times \cos e{c^2}x))dx} \]
Separating the integrals, we get,
\[\int {(1 - \cos x)\cos e{c^2}x} \] \[dx\]\[ = \int {\cos e{c^2}x} \] \[dx\] - \[\int {\cos x \times \cos e{c^2}x} \] \[dx\]
We know that multiplication of two minus signs yields a positive sign. Hence, we have.
\[\int {(1 - \cos x)\cos e{c^2}x} \] \[dx\]\[ = \int {( - )( - \cos e{c^2}x)dx} \]\[ - \int {\cos x \times \cos e{c^2}x} \] \[dx\]

As we know, \[(\cos ecx = \dfrac{1}{{\sin x}})\]. So, using this we have
\[\int {(1 - \cos x)\cos e{c^2}x} \] \[dx\]\[ = ( - )\int { - \cos ecx} \] \[dx\]\[ - \int {\cos x \times \dfrac{1}{{{{\sin }^2}x}}dx} \]
Using the formula \[\int { - \cos e{c^2}x = \cot x} + {c_1}\]
\[\int {(1 - \cos x)\cos e{c^2}x} \] \[dx\]\[ = ( - )(\cot x + {c_1}) - \int {\dfrac{{\cos x}}{{{{\sin }^2}x}}dx} \]
\[\Rightarrow \int {(1 - \cos x)\cos e{c^2}x} \] \[dx\]\[ = ( - )(\cot x + {c_1}) - \int {\dfrac{{\cos x}}{{{{\sin }^2}x}}dx} - - - - - - (1)\]
We will now solve \[\int {\dfrac{{\cos x}}{{{{\sin }^2}x}}dx} - - - - - - (2)\]

Let us suppose,
\[\sin x = t - - - - - - (3)\]\[\sin x = t - - - - - - (3)\]
Now, differentiating both sides, we get
\[\Rightarrow (\sin x)' = t'\]
\[\Rightarrow \cos x\] \[dx\] \[ = dt\]
Now, substituting \[\sin x = t\]and \[\cos x\]\[dx\]\[ = dt\]in (2), we get
\[\Rightarrow \int {\dfrac{{\cos x}}{{{{\sin }^2}x}}dx = \int {\dfrac{{dt}}{{{t^2}}}} } \]
\[\Rightarrow \int {{t^{ - 2}}dt} + {c_2}\]
Using the formula \[\int {{t^n}dt = \dfrac{{{t^{n + 1}}}}{{n + 1}}} + k\] for integration,
\[\Rightarrow \dfrac{{{t^{ - 1}}}}{{( - 1)}} + {c_2}\]
\[\Rightarrow ( - ){t^{ - 1}} + {c_2}\]
As \[{t^{ - 1}} = \dfrac{1}{t}\]. So, we can write the equation as
\[( - )\dfrac{1}{t} + {c_2}\]

Now, replacing back the value of t in the above equation, From (3) ,we get
\[\int {\dfrac{{\cos x}}{{{{\sin }^2}x}}dx = ( - )\dfrac{1}{{\sin x}} + {c_2}} \]
Substituting the above value in (1), we get,
\[\int {(1 - \cos x)\cos e{c^2}x} \]\[dx\]\[ = ( - )(\cot x + {c_1}) - (( - )\dfrac{1}{{\sin x}} + {c_2})\]
\[\Rightarrow \int {(1 - \cos x)\cos e{c^2}x} \]\[dx\]\[ = - \cot x - {c_1} + \dfrac{1}{{\sin x}} - {c_2}\]
Letting \[k = {c_1} + {c_2}\]
\[\Rightarrow \int {(1 - \cos x)\cos e{c^2}x} \]\[dx\]\[ = - \cot x + \dfrac{1}{{\sin x}} - k\]
\[\Rightarrow \int {(1 - \cos x)\cos e{c^2}x} \]\[dx\]\[ = - \cot x + \dfrac{1}{{\sin x}} - k - - - - - - (4)\]
Writing \[\cot x\] in terms of \[\sin x\]and \[\cos x\] in (4) i.e. \[\cot x = \dfrac{{\cos x}}{{\sin x}}\] ,we get
\[\int {(1 - \cos x)\cos e{c^2}x} \]\[dx\]\[ = - \dfrac{{\cos x}}{{\sin x}} + \dfrac{1}{{\sin x}} - k\]
\[\Rightarrow \int {(1 - \cos x)\cos e{c^2}x} \]\[dx\]\[ = \dfrac{{ - \cos x + 1}}{{\sin x}} - k - - - - - - (5)\]

Now, we are given the options in terms of \[\dfrac{x}{2}\]angle, so we need to transform our answer according to the given options.
We know,
\[1 - \cos 2t = 2{\sin ^2}t - - - - - - (6)\]
And,
\[\sin 2t = 2\sin t\cos t - - - - - - (7)\]
From (5), (6) and (7), Taking \[2t = x\],we get
\[\int {(1 - \cos x)\cos e{c^2}x} \]\[dx\]\[ = \dfrac{{2{{\sin }^2}\tfrac{x}{2}}}{{2\sin \tfrac{x}{2}\cos \tfrac{x}{2}}} - k\]
\[\Rightarrow \dfrac{{\sin \tfrac{x}{2}}}{{\cos \tfrac{x}{2}}} - k\]
Letting \[c = - k\]
\[\tan \tfrac{x}{2} + c\]
Hence, we got
\[\therefore \int {(1 - \cos x)\cos e{c^2}x} \]\[dx\]\[ = \tan \tfrac{x}{2} + c\]

Therefore, the correct option is A.

Note: The indefinite integrals of certain functions may have more than one answer in different forms. So, we have to solve in a particular way to reach the answer present in the options. However, all these forms are correct and interchangeable into one another. Indefinite integral gives us the family of curves as we don’t know the exact value of the arbitrary constant. The trigonometric formulas such as double angle formulae of sine and cosine are very useful in solving the given problem.