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If polynomials $P$ and $Q$ satisfies
$\begin{array}{l} \int ((3 x - 1) \cos x + (1 - 2 x) \sin x) d x \\ = P \cos x + Q \sin x \end{array}$
$(ignoring the constant of integration) then$

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P=3x−12

Q=2+x

P=3(x−1)

Q=3(x−1)

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detailed solution

Correct option is D

Integrating by parts, the given integral is equal to(3x−1)sin⁡x−3∫sin⁡xdx−(1−2x)cos⁡x−2∫cos⁡xdx=(3x−1)sin⁡x+3cos⁡x−(1−2x)cos⁡x−2sin⁡x=(3x−3)sin⁡x+(2+2x)cos⁡xHence Q=3(x−1) and P=2(1+x)

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If polynomials P and Q satisfies∫((3x−1)cos⁡x+(1−2x)sin⁡x)dx=Pcos⁡x+Qsin⁡x (ignoring the constant of integration) then

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

If polynomials $P$ and $Q$ satisfies
$\begin{array}{l} \int ((3 x - 1) \cos x + (1 - 2 x) \sin x) d x \\ = P \cos x + Q \sin x \end{array}$
$(ignoring the constant of integration) then$