Question

Solve $\int \left[(1-\cos x)\right]co{\sec }^{2}xdx=.....+c$

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

Answer 1

Answer: (A)

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

$I=\int (1-\cos x)co{\sec }^{2}xdx$
$=\int \frac{1-\cos x}{{\sin }^{2}x}dx=\int \frac{1-\cos x}{1-{\cos }^{2}x}dx$
$=\int \frac{1}{1+\cos x}dx=\int \frac{1}{2{\cos }^2\left(\frac{x}{2}\right)}dx$
$\therefore I=\int \frac{1}{2}{\sec }^2\left(\frac{x}{2}\right)=\int \tan tdt(\therefore \frac{x}{2}=t)$
$\therefore I=\tan \left(\frac{x}{2}\right)+c$