∫e2xsin⁡xdx is equal to

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ex5(sin⁡x−cos⁡x)+C

e2x5(2sin⁡x−cos⁡x)+C

e2x5(2cos⁡x−sin⁡x)+C

ex5(2cos⁡x−sin⁡x)+C

answer is B.

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Detailed Solution

Let, ∫e2xsin⁡xdxOn taking sin x as first function and e2x as second function and integrating by parts, we get∫e2xsin⁡xdx=sin⁡x⋅∫e2xdx−∫ddx(sin⁡x)⋅∫e2xdxdx=sin⁡x⋅e2x2−∫cos⁡x⋅e2x2dx=sin⁡x⋅e2x2−12∫e2xcos⁡xdx⇒∫e2xsin⁡xdx=sin⁡x⋅e2x2−12l1------iwhere I1=∫e2xcos⁡xdxOn taking cos x as first function and e2x as second function and integrating by parts, we getI1=cos⁡x⋅∫e2xdx−∫ddx(cos⁡x)⋅∫e2xdxdx=cos⁡x⋅e2x2−∫(−sin⁡x)⋅e2x2dx+C1=e2xcos⁡x2+12∫e2xsin⁡xdx+C1On putting value of I1 in Eq. (i), we get∫e2xsin⁡xdx=e2xsin⁡x2−12e2xcos⁡x2+12∫e2xsin⁡xdx+C1⇒∫e2xsin⁡xdx=e2xsin⁡x2−e2xcos⁡x4−14∫e2xsin⁡xdx−12C1⇒∫e2xsin⁡xdx+14∫e2xsin⁡xdx=e2xsin⁡x2−e2xcos⁡x4−12C1⇒ 54∫e2xsin⁡xdx=e2xsin⁡x2−e2xcos⁡x4−12C1⇒ ∫e2xsin⁡xdx=45e2xsin⁡x2−e2xcos⁡x4−12C1=25e2xsin⁡x−15e2xcos⁡x−25C1=25e2xsin⁡x−15e2xcos⁡x+Cwhere ,C=−25C1Hence , I=e2x5(2sin⁡x−cos⁡x+C)

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