Question

If ${\int }_0^{\alpha }\frac{dx}{1-cos\alpha cosx}=\frac{A}{sin\alpha }+B(\alpha \ne 0).$Then possible values of A and B are

• A. $A=\frac{\pi }{2},B=0$
• B. $A=\frac{\pi }{4},B=\frac{\pi }{4sin\alpha }$
  
• C. $A=\frac{\pi }{6},B=\frac{\pi }{sin\alpha }$
  
• D. $A=\pi ,B=\frac{\pi }{sin\alpha }$

Answer 1

Answer: (A), (B)

The correct options are
A

$A=\frac{\pi }{2},B=0$

B

$A=\frac{\pi }{4},B=\frac{\pi }{4sin\alpha }$

$I={\int }_0^{\alpha }\frac{dx}{1-cos\alpha cosx}$

$={\int }_0^{\alpha }\frac{dx}{2si{n}^2\left(\frac{\alpha }{2}\right)co{s}^2\frac{x}{2}+2co{s}^2\left(\frac{\alpha }{2}\right)si{n}^2\frac{x}{2}}$

$=\frac{1}{2}{\int }_0^{\alpha }\frac{\left(se{c}^2\frac{\alpha }{2}\right)se{c}^2\frac{x}{2}}{ta{n}^2\left(\frac{\alpha }{2}\right)+ta{n}^2\frac{x}{2}}dx$

Put $tan\frac{x}{2}=t,$ we have

$I={\int }_0^{tan\frac{\alpha }{2}}se{c}^2\left(\frac{\alpha }{2}\right)cot\left(\frac{\alpha }{2}\right)\left[ta{n}^{-1}\right(\frac{t}{tan\frac{\alpha }{2}}){]}_0^{tan\frac{\alpha }{2}}$

$=\frac{2}{sin\alpha }.\frac{\pi }{4}=\frac{\pi }{2sin\alpha }$

Thus, $\frac{A}{sin\alpha }+B=\frac{\pi }{2sin\alpha }$ satisfy the last equation.