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Question: Integrate \(\int{\left( 1-\cos x \right)\cos e{{c}^{2}}}x dx=\) \(\begin{aligned} &\left( A \rig...

Integrate ∫(1−cos⁡x)cos⁡ec2xdx=\int{\left( 1-\cos x \right)\cos e{{c}^{2}}}x dx=
(A)−tan⁡x2+c       (B)tan⁡x2+c  (C)−2tan⁡x2+c    (D)2tan⁡x2+c \begin{aligned} &\left( A \right)-\tan \dfrac{x}{2}+c\,\,\,\,\,\,\,\left( B \right)\tan \dfrac{x}{2}+c\, \\\ &\left( C \right)-2\tan \dfrac{x}{2}+c\,\,\,\,\left( D \right)2\tan \dfrac{x}{2}+c \\\ \end{aligned}

Explanation

Solution

For this question, first open the bracket, then separate it into two parts. Suppose individual parts in the form of variables like m, n and at last apply integration formulae.

Complete step by step solution:
The given integrand is
∫(1−cos⁡x)cos⁡ec2xdx\int{\left( 1-\cos x \right)\cos e{{c}^{2}}}x dx bracket with
First we will open the brackets with help of multiplication then we have to divide it into two terms or parts and integrate individual parts , so that
∫(1−cos⁡x)cos⁡ec2xdx=\int{\left( 1-\cos x \right)\cos e{{c}^{2}}}x dx= ∫cos⁡ec2xdx−∫cos⁡x.cos⁡ec2xdx\int{\cos e{{c}^{2}}}x dx-\int{\cos x.\cos e{{c}^{2}}}x dx……………………..(i)
Now to integrate it, let us suppose sin⁡x=m\sin x= m
If we differentiate sin⁡x=m\sin x=m, we get by the differentiation formula,
ddxsin⁡x=dm →cos⁡xdx=dm \begin{aligned} & \dfrac{d}{dx}\sin x=dm \\\ & \rightarrow \cos x dx=dm \\\ \end{aligned}
We know that cos⁡ec2xdx=−cot⁡x\cos e{{c}^{2}}x dx=-\cot x
On putting these values in equation (i)
∫cos⁡ec2xdx−∫cos⁡x.cos⁡ec2xdx\int{\cos e{{c}^{2}}}xdx-\int{\cos x.\cos e{{c}^{2}}}xdx=−cot⁡x−∫dmm2-\cot x-\int{\dfrac{dm}{{{m}^{2}}}}
By integration formula, we know that ∫xndx=xn+1n+1+c\int{{{x}^{n}}}dx=\dfrac{{{x}^{n+1}}}{n+1}+c, where c is constant of integration.
∫cos⁡ec2xdx−∫cos⁡x.cos⁡ec2xdx\int{\cos e{{c}^{2}}}xdx-\int{\cos x.\cos e{{c}^{2}}}xdx =−cot⁡x+1m-\cot x+\dfrac{1}{m}
∫cos⁡ec2xdx−∫cos⁡x.cos⁡ec2xdx\int{\cos e{{c}^{2}}}xdx-\int{\cos x.\cos e{{c}^{2}}}xdx =−cot⁡x+1sin⁡x-\cot x+\dfrac{1}{\sin x}
Because we already supposed that sin⁡x=m\sin x=m
Again we can write cot⁡x=cos⁡xsin⁡x\cot x=\dfrac{\cos x}{\sin x}, then
∫cos⁡ec2xdx−∫cos⁡x.cos⁡ec2xdx\int{\cos e{{c}^{2}}}xdx-\int{\cos x.\cos e{{c}^{2}}}xdx=−cos⁡xsin⁡x+1sin⁡x-\dfrac{\cos x}{\sin x}+\dfrac{1}{\sin x}
By common denominator rule of subtraction,
∫cos⁡ec2xdx−∫cos⁡x.cos⁡ec2xdx\int{\cos e{{c}^{2}}}xdx-\int{\cos x.\cos e{{c}^{2}}}xdx=−cos⁡x+1sin⁡x=1−cos⁡xsin⁡x-\dfrac{\cos x+1}{\sin x}=\dfrac{1-\cos x}{\sin x}
By the identity of trigonometry,
∫cos⁡ec2xdx−∫cos⁡x.cos⁡ec2xdx\int{\cos e{{c}^{2}}}xdx-\int{\cos x.\cos e{{c}^{2}}}xdx=2sin⁡2x22sin⁡x2cos⁡x2\dfrac{2{{\sin }^{2}}\dfrac{x}{2}}{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}
sin⁡x/2  cancel out to sin⁡x/2\sin x/2\,\,cancel\,out\,to\,\sin x/2
∫cos⁡ec2xdx−∫cos⁡x.cos⁡ec2xdx\int{\cos e{{c}^{2}}}xdx-\int{\cos x.\cos e{{c}^{2}}}xdx=sin⁡x2cos⁡x2\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}
We know thatsin⁡x2cos⁡x2=tan⁡x2\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}=\tan \dfrac{x}{2}, then we get
∫cos⁡ec2xdx−∫cos⁡x.cos⁡ec2xdx\int{\cos e{{c}^{2}}}xdx-\int{\cos x.\cos e{{c}^{2}}}xdx=tan⁡x2+c\tan \dfrac{x}{2}+c
Where c is constant of integration.

Hence, The option (B) is the correct option.

Additional information: If more than one constant of integration is used while solving the integral, then at the end of the solution write only one constant of integration.
If the denominators of two fractions when they are in addition or subtraction, then we don’t need to take LCM .

Note: Sometimes students make mistakes, they apply the integration formula instead of before opening the brackets, then their solution would be wrong. It's important to solve such types of questions with the help of formulae and identities.