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Integral sin^2 alpha - sin^x divided by cos x - cos alpha = f(x)+Ax+B

Integral sin^2 alpha - sin^x divided by cos x - cos alpha = f(x)+Ax+B

Grade:12

2 Answers

Deepak Kumar Shringi
askIITians Faculty 4404 Points
5 years ago
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nithya
17 Points
5 years ago
ans:f(x)=sinx,A=cos (alpha)
 
in this way we can get the answer;
 
formulas used 
 
a^2-b^2=(a-b)(a+b)
 
sinA+sinB=2sin(A+B/2)cos(A-B/2)
 
 
 
\int sin^2\alpha -sin^2x/cosx-cos\alpha dx =\int \left [ 2sin(\alpha -x/2)cos(\alpha +x/2)*2sin(\alpha +x/2)cos(\alpha -x/2)\right ] dx/\left [ 2sin(x+\alpha /2)sin(\alpha -x /2)\ \right ] = \int 2cos(\alpha +x/2)cos(\alpha -x/2)dx=\int \left ( cos\alpha +cosx \right )dx=\left ( xcos\alpha +sinx \right ) \Rightarrow f(x=sinx),A=cos\alpha

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