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Question
`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to ______.
Options
2(sinx + xcosθ) + C
2(sinx – xcosθ) + C
2(sinx + 2xcosθ) + C
2(sinx – 2x cosθ) + C
Solution
`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to 2(sinx + xcosθ) + C.
Explanation:
Let I = `int (cos2x - cos 2theta)/(cosx - costheta) "d"x`
= `int ((2cos^2x - 1 - 2 cos^2theta + 1))/(cosx - costheta) "d"x`
= `2int ((cosx + cos theta)(cosx - costheta))/((cosx - costheta)) "d"x`
= `2int(cos x + cos theta) "d"x`
= 2(sinx + xcosθ) + C
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