Question

Integrate the given expression, \[\int{\left( 1-cosx \right)cose{{c}^{2}}x.dx}\] and find \[\operatorname{f}(x)\] , if \[\int{\left( 1-cosx \right)cose{{c}^{2}}x.dx}=f(x)+C\] .

Answer 1

Hint: Expand the given expression as \[\int{cose{{c}^{2}}x.dx-\int{cosx.cose{{c}^{2}}x.dx}}\] . Integrate both terms of the expression separately and We know that \[\int{cose{{c}^{2}}x.dx}=-\cot x\] . Then replace \[{{\operatorname{cosec}}^{2}}x\] by \[\dfrac{1}{si{{n}^{2}}x}\] in the term \[\int{cosx.cose{{c}^{2}}x.dx}\] . Assume \[\operatorname{t}=sinx\] . Then, replace \[\operatorname{cosxdx}\] as \[\operatorname{dt}\] . Now, integrate the expression \[\int{\dfrac{1}{{{t}^{2}}}dt}\] and solve further.


Complete step-by-step solution -
According to the question, we have to integrate the expression \[\int{\left( 1-cosx \right)cose{{c}^{2}}x.dx}\] ………..(1)
To integrate this expression, first of all, we have to convert this expression into a simpler form.
Now, expanding equation (1), we get
\[\int{\left( 1-cosx \right)cose{{c}^{2}}x.dx}\]
\[=\int{\left( cose{{c}^{2}}x-cosx.cose{{c}^{2}}x \right).dx}\]
\[=\int{cose{{c}^{2}}x.dx-\int{cosx.cose{{c}^{2}}x.dx}}\] ………………..(2)
We know that, \[\int{cose{{c}^{2}}x.dx}=-\cot x\] ……………………(3)
Using equation (3), we can transform equation (2).
\[\begin{align}
  & \int{\left( cose{{c}^{2}}x-cosx.cose{{c}^{2}}x \right).dx} \\
 & =\int{cose{{c}^{2}}x.dx-\int{cosx.cose{{c}^{2}}x.dx}} \\
\end{align}\]
\[=-cotx-\int{cosx.cose{{c}^{2}}x.dx}\] ……………………..(4)
Now, we have to integrate the expression, \[\int{\cos x.\cos e{{c}^{2}}x.dx}\] …………………(5)
Replacing, \[{{\operatorname{cosec}}^{2}}x\] by \[\dfrac{1}{si{{n}^{2}}x}\] in the equation (5), we get
\[\begin{align}
  & \int{cosx.cose{{c}^{2}}x.dx} \\
 & =\int{cosx.\dfrac{1}{si{{n}^{2}}x}.dx} \\
\end{align}\]
\[=\int{\dfrac{cosx}{sinx.sinx}dx}\] …………………………(6)
Let us assume \[\operatorname{t}=sinx\] ……………….(7)
Differentiating with respect to x in equation (7), we get
\[\dfrac{dt}{dx}=cosx\]
\[\Rightarrow dt=cosx.dx\] ………………(8)
Now, integrating equation (8), we get
\[\Rightarrow \int{dt}=\int{cosx.dx}\] …………………………….(9)
Now, using equation (7), we can transform equation (6).
Transforming equation (6),
\[=\int{\dfrac{cosx}{t.t}dx}\] …………….(10)
Now, using equation (9), we can write equation (10) as
\[=\int{\dfrac{cosx}{t.t}dx}\]
\[=\int{\dfrac{1}{{{t}^{2}}}dt}\] …………………….(11)
Now, integrating the above equation (11)
\[=\int{\dfrac{1}{{{t}^{2}}}dt}\]
\[=\dfrac{{{t}^{-2+1}}}{-2+1}\]
\[=\dfrac{{{t}^{-1}}}{-1}\] ……………….(12)
Now, using equation (7), we can write equation (12) as,
\[=\dfrac{{{t}^{-1}}}{-1}\]
\[=-\dfrac{1}{sinx}\] ………………….(13)
We have got \[\int{\dfrac{cosx}{sinx.sinx}dx}=-\dfrac{1}{sinx}\] .
Now, we can write equation (4) as
\[\begin{align}
  & =-cotx-\int{cosx.cose{{c}^{2}}x.dx} \\
 & =-cotx-\left( -\dfrac{1}{sinx} \right) \\
\end{align}\]
We know that, \[\operatorname{cotx}=\dfrac{cosx}{sinx}\] .
\[\begin{align}
  & -cotx+sinx+C \\
 & =-\dfrac{cosx}{sinx}+\dfrac{1}{sinx} \\
\end{align}\]
\[=\dfrac{(1-cosx)}{\sin x}\] ………………..(14)
We know that, \[\operatorname{sinx}=2sin\dfrac{x}{2}.cos\dfrac{x}{2}\] ……………………..(15)
We also know that,
\[cosx=1-2si{{n}^{2}}\dfrac{x}{2}\]
\[\Rightarrow 2si{{n}^{2}}\dfrac{x}{2}=1-cosx\] ……………………(16)
Now, using equation (15) and equation (16) in equation (14), we get
\[\begin{align}
  & =\dfrac{(1-cosx)}{\sin x} \\
 & =\dfrac{2{{\sin }^{2}}\dfrac{x}{2}}{2\sin \dfrac{x}{2}.cos\dfrac{x}{2}} \\
\end{align}\]
\[\begin{align}
  & =\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}} \\
 & =\tan \dfrac{x}{2} \\
\end{align}\]
So, \[\int{\left( 1-cosx \right)cose{{c}^{2}}x.dx}=tan\dfrac{x}{2}+C\] , where C is a constant………..(17)
According to the question, we have \[\int{\left( 1-cosx \right)cose{{c}^{2}}x.dx}=f(x)+C\] …………….(18)
Now, comparing equation (17) and equation (18), we get
\[\begin{align}
  & \operatorname{f}(x)+C=tan\dfrac{x}{2}+C \\
 & \Rightarrow f(x)=tan\dfrac{x}{2} \\
\end{align}\]
Hence, \[f(x)=tan\dfrac{x}{2}\] .

Note: In this question, one can think to assume \[\operatorname{cosx}\] as t in the expression \[\int{cosx.cose{{c}^{2}}x.dx}\] . But, if we do so then our equation will look like \[\int{t.\dfrac{1}{si{{n}^{2}}x}dx}\] . We can see that this expression has become complex to be solved further. So, this approach is not suitable for the integration of this expression.