∫0α dx1−cos⁡xcos⁡α=Asin⁡α+B(α≠0)possible value of A and B are

I=∫0α dx1−cos⁡α⋅cos⁡x =∫0α dxcos2⁡x2+sin2⁡x2−cos⁡αcos2⁡x2−sin2⁡x2 =∫0α dx(1−cos⁡α)cos2⁡x2+(1+cos⁡α)sin2⁡x2=∫0α dx2sin2⁡α2cos2⁡x2+2cos2⁡α2sin2⁡x2=12∫0α sec2⁡α2sec2⁡x2tan2⁡α2+tan2⁡x2dx Put tanx/2 = t, we have I=∫0tan⁡α/2 sec2⁡α2t2+tan2⁡α2dt =sec2⁡α2cot⁡α2tan−1⁡ttan⁡α20tan⁡α2 =2sin⁡α⋅π4=π2sin⁡α Asin⁡α+B=π2sin⁡α ⇒A=π2,B=0 or A=π4,B=π4sin⁡α

∫0α dx1−cos⁡xcos⁡α=Asin⁡α+B(α≠0)possible value of A and B are

I=∫0α dx1−cos⁡α⋅cos⁡x =∫0α dxcos2⁡x2+sin2⁡x2−cos⁡αcos2⁡x2−sin2⁡x2 =∫0α dx(1−cos⁡α)cos2⁡x2+(1+cos⁡α)sin2⁡x2=∫0α dx2sin2⁡α2cos2⁡x2+2cos2⁡α2sin2⁡x2=12∫0α sec2⁡α2sec2⁡x2tan2⁡α2+tan2⁡x2dx Put tanx/2 = t, we have I=∫0tan⁡α/2 sec2⁡α2t2+tan2⁡α2dt =sec2⁡α2cot⁡α2tan−1⁡ttan⁡α20tan⁡α2 =2sin⁡α⋅π4=π2sin⁡α Asin⁡α+B=π2sin⁡α ⇒A=π2,B=0 or A=π4,B=π4sin⁡α

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$\int_{0}^{α} \frac{d x}{1 - \cos x \cos α} = \frac{A}{\sin α} + B (α \neq 0)$ possible value of A and B are

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$A = \frac{π}{2}, B = 0$

$A = \frac{π}{4}, B = \frac{π}{4 \sin α}$

$A = \frac{π}{6}, B = \frac{π}{\sin α}$

$A = π, B = \frac{π}{\sin α}$

answer is A, B.

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Detailed Solution

$I = \int_{0}^{α} \frac{dx}{1 - \cos α \cdot \cos x} = \int_{0}^{α} \frac{dx}{(\cos^{2} \frac{x}{2} + \sin^{2} \frac{x}{2}) - \cos α (\cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2})} \begin{array}{l} = \int_{0}^{α} \frac{d x}{(1 - \cos α) \cos^{2} \frac{x}{2} + (1 + \cos α) \sin^{2} \frac{x}{2}} \\ = \int_{0}^{α} \frac{d x}{2 \sin^{2} \frac{α}{2} \cos^{2} \frac{x}{2} + 2 \cos^{2} \frac{α}{2} \sin^{2} \frac{x}{2}} \\ = \frac{1}{2} \int_{0}^{α} \frac{\sec^{2} \frac{α}{2} \sec^{2} \frac{x}{2}}{\tan^{2} \frac{α}{2} + \tan^{2} \frac{x}{2}} d x \end{array} Put tanx / 2 = t, we have I = \int_{0}^{\tan α / 2} \frac{\sec^{2} \frac{α}{2}}{t^{2} + \tan^{2} \frac{α}{2}} dt = \sec^{2} \frac{α}{2} \cot \frac{α}{2} {|\tan^{- 1} (\frac{t}{\tan \frac{α}{2}})|}_{0}^{\tan \frac{α}{2}} = \frac{2}{\sin α} \cdot \frac{π}{4} = \frac{π}{2 \sin α} \frac{A}{\sin α} + B = \frac{π}{2 \sin α} \Rightarrow A = \frac{π}{2}, B = 0 or A = \frac{π}{4}, B = \frac{π}{4 \sin α}$