Find: ∫ex.sin2xdx - Mathematics

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Sum

Find: `int e^x.sin2xdx`

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Solution

Let I = `int e^xsin2xdx`

Applying integration by parts

I = `int \underset(\text(I))(e)^x \underset(\text(II))(sin 2x) dx`

= `e^x int sin 2xdx - int [d/(dx) (e^x) int sin 2xdx]dx`

= `e^x((-cos2x)/2) + 1/2 int e^x cos 2xdx`

= `1/2(-e^x cos2x) + 1/2[e^x int cos 2xdx - int (d/(dx) (e^x) int cos2xdx)dx]`

= `1/2 (-e^x cos2x) + 1/2[(e^xsin2x)/2 - 1/2 int e^x sin 2xdx]`

= `1/2 (-e^x cos 2x) + 1/4 (e^x sin 2x) - 1/4 int e^x sin 2xdx + K`

∴ 4I = `-2e^x cos2x + e^xsin2x - I + K`

or 5I = `-2e^x cos2x + e^xsin2x + K`

I = `1/5(e^xsin2x - 2e^xcos2x) + K/5`

or I = `1/5(e^xsin2x - 2e^xcos2x) + c`

  Is there an error in this question or solution?
2021-2022 (April) Delhi Set 1

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