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).

& \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{aligned}$$ $$=-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{aligned} & \int{cosx.cose{{c}^{2}}x.dx} \\\ & =\int{cosx.\dfrac{1}{si{{n}^{2}}x}.dx} \\\ \end{aligned}$$ $$=\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{aligned} & =-cotx-\int{cosx.cose{{c}^{2}}x.dx} \\\ & =-cotx-\left( -\dfrac{1}{sinx} \right) \\\ \end{aligned}$$ We know that, $$\operatorname{cotx}=\dfrac{cosx}{sinx}$$ . $$\begin{aligned} & -cotx+sinx+C \\\ & =-\dfrac{cosx}{sinx}+\dfrac{1}{sinx} \\\ \end{aligned}$$ $$=\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{aligned} & =\dfrac{(1-cosx)}{\sin x} \\\ & =\dfrac{2{{\sin }^{2}}\dfrac{x}{2}}{2\sin \dfrac{x}{2}.cos\dfrac{x}{2}} \\\ \end{aligned}$$ $$\begin{aligned} & =\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}} \\\ & =\tan \dfrac{x}{2} \\\ \end{aligned}$$ So, $$\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{aligned} & \operatorname{f}(x)+C=tan\dfrac{x}{2}+C \\\ & \Rightarrow f(x)=tan\dfrac{x}{2} \\\ \end{aligned}$$ Hence, $$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.