Question

$\int (1-\cos x)\cos e{c}^{2}xdx=f\left(x\right)+c\Rightarrow f\left(x\right)=$

• A. $\tan \frac{x}{2}$
• B. $cot\frac{x}{2}$
• C. $2\tan \frac{x}{2}$
• D. $\frac{1}{2}\tan \frac{x}{2}$

Answer 1

The correct option is A $\tan \frac{x}{2}$

$\int (1-\cos x)co{\sec }^{2}xdx$

Let $u=1-\cos x\therefore du=\sin x;v=co{\sec }^{2}x\therefore v=-\cot x$

Now integrating by parts, we get

$(1-\cos x)\int co{\sec }^{2}x-\int \sin x.\int cose{c}^{2}xdx=(1-\cos x)\left(\cot x\right)+\int \sin x.\cot xdx-\cot x(1-\cos x)+\sin x+c$

$\Rightarrow \frac{{\cos }^{2}x-\cos x+{\sin }^{2}x}{\sin x}+c=\frac{1-\cos x}{\sin x}+c\Rightarrow \tan \frac{x}{2}+c$

$\Rightarrow f\left(x\right)=\tan \left(\frac{x}{2}\right)$