Question

Integrate $\dfrac{\left[ x-\sin x \right]}{\left[ 1-\cos x \right]}$

Answer 1

In this type of question we have to use the concept of trigonometry. Here, we have to use trigonometric formulas such as $\sin 2\theta =2\sin \theta \cos \theta$ and $\cos 2\theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta =1-2{{\sin }^{2}}\theta =2{{\cos }^{2}}\theta -1$. Also we have to make some adjustments depending on the formulas which we have used for simplification of the trigonometric expression specifically when we express $\sin x$ and $\cos x$ by using the formulas of $\sin 2\theta$ and $\cos 2\theta$. We will use integration by parts for evaluating the integral. We know that the formula for integration by parts is given by $\int{uvdx=u\int{vdx-\int{\left( \dfrac{d}{dx}u\centerdot \int{vdx} \right)dx}}}$.

Complete step-by-step solution:
Now in this question we have to integrate the function $\dfrac{\left[ x-\sin x \right]}{\left[ 1-\cos x \right]}$. So that let us consider,
$\Rightarrow I=\int{\dfrac{\left[ x-\sin x \right]}{\left[ 1-\cos x \right]}dx}$
As we know that, $\sin 2\theta =2\sin \theta \cos \theta$ hence we can express $\sin x$ as $\sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$.
Also as we have $\cos 2\theta =1-2{{\sin }^{2}}\theta$ we can express $\cos x$ as $\cos x=1-2{{\sin }^{2}}\dfrac{x}{2}$.
Hence, the above integral becomes,
$\Rightarrow I=\int{\dfrac{\left[ x-2\sin \dfrac{x}{2}\cos \dfrac{x}{2} \right]}{\left[ 1-\left( 1-2{{\sin }^{2}}\dfrac{x}{2} \right) \right]}}dx$
$\Rightarrow I=\int{\dfrac{\left[ x-2\sin \dfrac{x}{2}\cos \dfrac{x}{2} \right]}{\left[ 2{{\sin }^{2}}\dfrac{x}{2} \right]}}dx$
By separating the denominator over the subtraction present in the numerator we get,
$\Rightarrow I=\int{\left[ \dfrac{x}{2{{\sin }^{2}}\dfrac{x}{2}}-\dfrac{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}{2{{\sin }^{2}}\dfrac{x}{2}} \right]}dx$
$\Rightarrow I=\dfrac{1}{2}\int{\dfrac{x}{{{\sin }^{2}}\dfrac{x}{2}}dx}-\int{\dfrac{\cos \dfrac{x}{2}}{\sin \dfrac{x}{2}}dx}$
As we know that, $\dfrac{1}{\sin \theta }=\text{cosec}\theta ,\dfrac{\cos \theta }{\sin \theta }=\cot \theta$ . By using this we get,
$\Rightarrow I =\dfrac{1}{2}\int{x\cos e{{c}^{2}}\dfrac{x}{2}dx-\int{\cot \dfrac{x}{2}dx}}$
By using integration by parts in first integral and keeping the second integral as it is we can write,
$\Rightarrow I=\dfrac{1}{2}\left[ x\int{\cos e{{c}^{2}}\dfrac{x}{2}dx-\int{\left( \dfrac{d}{dx}x\int{\cos e{{c}^{2}}\dfrac{x}{2}dx} \right)dx}} \right]-\int{\cot \dfrac{x}{2}dx}$
Now, as we know that, $\int{\cos e{{c}^{2}}x=-\cot x+C}$ we can write above integral as,
$\Rightarrow I=\dfrac{1}{2}\left[ x\left( -2\cot \dfrac{x}{2} \right)-\int{\left( -2\cot \dfrac{x}{2} \right)}dx \right]-\int{\cot \dfrac{x}{2}dx}$
$\Rightarrow I=\dfrac{1}{2}\left( -2x\cot \dfrac{x}{2} \right)+2\int{\cot \dfrac{x}{2}dx}-\int{\cot \dfrac{x}{2}dx}$
$\Rightarrow I=-x\cot \dfrac{x}{2}+\int{\cot \dfrac{x}{2}dx}$
$\Rightarrow I=-x\cot \dfrac{x}{2}+2\ln \sin \dfrac{x}{2}+C$
Hence, $\int{\dfrac{\left[ x-\sin x \right]}{\left[ 1-\cos x \right]}dx}=-x\cot \dfrac{x}{2}+2\ln \sin \dfrac{x}{2}+C$

Note: In this type of question students have to remember the formula of integration by parts. Also as in such questions students have to use the formulas of integration of different trigonometric functions so they have to remember formulas of integration of all trigonometric functions. Also students have to take care in integrating the trigonometric functions having angles such as $\dfrac{x}{2}$.