Question

$\int (1-\cos x)\cos e{c}^{2}xdx=$

• A. $\tan \left(\frac{x}{2}\right)+c$
• B. $–cot\left(\frac{x}{2}\right)+c$
• C. $2\tan \left(\frac{x}{2}\right)+c$
• D. $–2cot\left(\frac{x}{2}\right)+c$

Answer 1

The correct option is A

$\tan \left(\frac{x}{2}\right)+c$

Explanation for the correct answer:

Given

$$\begin{gathered} \int (1-\cos x)\cos e{c}^{2}xdx=\int \frac{(1–\cos x)}{{\sin }^{2}x}dx \\ =\int \frac{(1–\cos x)}{(1–{\cos }^{2}x)}dx\left[\because {\sin }^{2}x=1–{\cos }^{2}x\right] \\ =\int \frac{(1–\cos x)}{(1+\cos x)(1-\cos x)}dx\left[\because \left(a^2-b^2\right)=\left(a+b\right)\left(a-b\right)\right] \\ =\int \frac{1}{(1+\cos x)}dx \\ =\int \frac{dx}{1+2{\cos }^2{\displaystyle \left(\frac{x}{2}\right)}–1}\left[\because \cos x=2{\cos }^2\left(\frac{x}{2}\right)–1\right] \\ =\int se{c}^2\left(\frac{x}{2}\right)dx \\ I=\tan \left(\frac{x}{2}\right)+c\left[\because \int se{c}^2\left(\frac{x}{2}\right)dx\right] \end{gathered}$$

Therefore, $\int (1-\cos x)\cos e{c}^{2}xdx=$$\tan \left(\frac{x}{2}\right)+c$

Hence, the correct answer is option (A).