Question

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

• A. $A=\frac{\pi }{2},B=0$
• B. $A=\frac{\pi }{8},B=\frac{\pi }{4sin\alpha }$
  
• C. $A=\frac{\pi }{6},B=\frac{\pi }{sin\alpha }$
  
  
• D. $A=\pi ,B=\frac{\pi }{sin\alpha }$

Answer 1

The correct option is A

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

$I={\int }_0^{\alpha }\frac{dx}{1-\cos \alpha \cos x}={\int }_0^{\alpha }\frac{dx}{co{s}^2\frac{x}{2}+si{n}^2\frac{x}{2}-cos\alpha (co{s}^2\frac{x}{2}-si{n}^2\frac{x}{2})}={\int }_0^{\alpha }\frac{dx}{(1-cos\alpha )co{s}^2\frac{x}{2}+(1+cos\alpha )si{n}^2\frac{x}{2}}={\int }_0^{\alpha }\frac{se{c}^2\frac{x}{2}dx}{2si{n}^2\frac{\alpha }{2}+2co{s}^2\frac{\alpha }{2}ta{n}^2\frac{x}{2}}=\frac{1}{2}{\int }_0^{\alpha }\frac{se{c}^2\frac{\alpha }{2}se{c}^2\frac{x}{2}dx}{ta{n}^2\frac{\alpha }{2}+ta{n}^2\frac{x}{2}}$
Let $tan\frac{x}{2}=tI={\int }_0^{tan\alpha /2}\frac{se{c}^2\frac{\alpha }{2}dt}{t^2+ta{n}^2\frac{\alpha }{2}}=se{c}^2\frac{\alpha }{2}cot\frac{\alpha }{2}{\left[ta{n}^{-1}\left(\frac{t}{tan\frac{\alpha }{2}}\right)\right]}_0^{tan\frac{\alpha }{2}}=\frac{2}{sin\alpha }.\frac{\pi }{4}=\frac{\pi }{2sin\alpha }Now\frac{A}{sin\alpha }+B=\frac{\pi }{2sin\alpha }\Rightarrow A=\frac{\pi }{2},B=0alsoA=\frac{\pi }{4},B=\frac{\pi }{4sin\alpha }$
Hence (A), (B) are correct answer.