The range of the function f(x)=∫−11sinx1−2tcosx+t2dt

∫−11 sin⁡x1−cos2⁡x+cos2⁡x−2tcos⁡x+t2dx∫−11 sin⁡xdxsin2⁡x+(t−cos⁡x)2(sin⁡x)⋅1sin⁡xtan−1⁡t−cos⁡xsin⁡x−11=tan−1⁡1−cos⁡xsin⁡x+tan−1⁡1+cos⁡xsin⁡x=tan−1⁡tan⁡x2+tan−1⁡1tan⁡x2=π2 if tan⁡x2>0=−π2 if tan⁡x2<0

The range of the function f(x)=∫−11sinx1−2tcosx+t2dt

∫−11 sin⁡x1−cos2⁡x+cos2⁡x−2tcos⁡x+t2dx∫−11 sin⁡xdxsin2⁡x+(t−cos⁡x)2(sin⁡x)⋅1sin⁡xtan−1⁡t−cos⁡xsin⁡x−11=tan−1⁡1−cos⁡xsin⁡x+tan−1⁡1+cos⁡xsin⁡x=tan−1⁡tan⁡x2+tan−1⁡1tan⁡x2=π2 if tan⁡x2>0=−π2 if tan⁡x2<0

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The range of the function $f (x) = \int_{- 1}^{1} \frac{sinx}{1 - 2 tcosx + t^{2}} dt$

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${0, π}$

$[0, π]$

$[\frac{- π}{2}, \frac{π}{2}]$

${\frac{- π}{2}, \frac{π}{2}}$

answer is D.

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Detailed Solution

$\begin{array}{l} \int_{- 1}^{1} \frac{\sin x}{1 - \cos^{2} x + \cos^{2} x - 2 tcos x + t^{2}} dx \\ \int_{- 1}^{1} \frac{\sin xdx}{\sin^{2} x + (t - \cos x)^{2}} \\ (\sin x) \cdot \frac{1}{\sin x} \tan^{- 1} {(\frac{t - \cos x}{\sin x})}_{- 1}^{1} \\ = \tan^{- 1} (\frac{1 - \cos x}{\sin x}) + \tan^{- 1} (\frac{1 + \cos x}{\sin x}) \\ = \tan^{- 1} (\tan \frac{x}{2}) + \tan^{- 1} (\frac{1}{\tan \frac{x}{2}}) \\ = \frac{π}{2} if \tan \frac{x}{2} > 0 \\ = \frac{- π}{2} if \tan \frac{x}{2} < 0 \end{array}$