Answer:
[tex]\qquad\boxed{ \sf{ \:\bf \: (f(x), \: A) = (sinx, \: cos \alpha ) \: }}\\ \\ [/tex]
Step-by-step explanation:
Given that,
[tex]\sf \:\displaystyle\int\sf \dfrac{ {sin}^{2} \alpha - {sin}^{2}x }{cosx - cos \alpha } \: dx = f(x) + Ax + B \\ \\ [/tex]
can be rewritten as
[tex]\sf \:\displaystyle\int\sf \dfrac{(1 - {cos}^{2} \alpha) - (1 - {cos}^{2}x) }{cosx - cos \alpha } \: dx = f(x) + Ax + B \\ \\ [/tex]
[tex]\sf \:\displaystyle\int\sf \dfrac{1 - {cos}^{2} \alpha - 1 + {cos}^{2}x}{cosx - cos \alpha } \: dx = f(x) + Ax + B \\ \\ [/tex]
[tex]\sf \:\displaystyle\int\sf \dfrac{ - {cos}^{2} \alpha + {cos}^{2}x}{cosx - cos \alpha } \: dx = f(x) + Ax + B \\ \\ [/tex]
[tex]\sf \:\displaystyle\int\sf \dfrac{ {cos}^{2}x - {cos}^{2} \alpha }{cosx - cos \alpha } \: dx = f(x) + Ax + B \\ \\ [/tex]
[tex]\sf \:\displaystyle\int\sf \dfrac{(cosx + cos \alpha )(cosx - cos \alpha )}{cosx - cos \alpha } \: dx = f(x) + Ax + B \\ \\ [/tex]
[tex]\sf \:\displaystyle\int\sf (cosx + cos \alpha ) \: dx = f(x) + Ax + B \\ \\ [/tex]
[tex]\sf \:sinx + x \: cos \alpha + B = f(x) + Ax + B \\ \\ [/tex]
So, on comparing, we get
[tex]\sf \:f(x) = sinx \\ \\ [/tex]
[tex]\sf \:A = cos \alpha \\ \\ [/tex]
Hence,
[tex]\bf\implies \: (f(x), \: A) = (sinx, \: cos \alpha ) \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Formulae Used
[tex]\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \: f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}[/tex]
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Answers & Comments
Verified answer
Answer:
[tex]\qquad\boxed{ \sf{ \:\bf \: (f(x), \: A) = (sinx, \: cos \alpha ) \: }}\\ \\ [/tex]
Step-by-step explanation:
Given that,
[tex]\sf \:\displaystyle\int\sf \dfrac{ {sin}^{2} \alpha - {sin}^{2}x }{cosx - cos \alpha } \: dx = f(x) + Ax + B \\ \\ [/tex]
can be rewritten as
[tex]\sf \:\displaystyle\int\sf \dfrac{(1 - {cos}^{2} \alpha) - (1 - {cos}^{2}x) }{cosx - cos \alpha } \: dx = f(x) + Ax + B \\ \\ [/tex]
[tex]\sf \:\displaystyle\int\sf \dfrac{1 - {cos}^{2} \alpha - 1 + {cos}^{2}x}{cosx - cos \alpha } \: dx = f(x) + Ax + B \\ \\ [/tex]
[tex]\sf \:\displaystyle\int\sf \dfrac{ - {cos}^{2} \alpha + {cos}^{2}x}{cosx - cos \alpha } \: dx = f(x) + Ax + B \\ \\ [/tex]
[tex]\sf \:\displaystyle\int\sf \dfrac{ {cos}^{2}x - {cos}^{2} \alpha }{cosx - cos \alpha } \: dx = f(x) + Ax + B \\ \\ [/tex]
[tex]\sf \:\displaystyle\int\sf \dfrac{(cosx + cos \alpha )(cosx - cos \alpha )}{cosx - cos \alpha } \: dx = f(x) + Ax + B \\ \\ [/tex]
[tex]\sf \:\displaystyle\int\sf (cosx + cos \alpha ) \: dx = f(x) + Ax + B \\ \\ [/tex]
[tex]\sf \:sinx + x \: cos \alpha + B = f(x) + Ax + B \\ \\ [/tex]
So, on comparing, we get
[tex]\sf \:f(x) = sinx \\ \\ [/tex]
[tex]\sf \:A = cos \alpha \\ \\ [/tex]
Hence,
[tex]\bf\implies \: (f(x), \: A) = (sinx, \: cos \alpha ) \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Formulae Used
[tex]\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \: f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}[/tex]