Integration by Parts with exponential and trigonometric

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TEST PREP QUESTION Case 5: Exponential and Trigonometric

$\int e^x \sin x \, dx = \int u \, dv = uv - \int v \, du$

This is likely to be on the exam. It is not hard to do. But if you do not know the trick, it is impossible. And the trick is just basic arithmetic logic.

$u_1 = e^x$ $du_1 = e^x dx$

$dv_1 = \sin x \, dx$ $v_1 = -\cos x$

$\int e^x \sin x \, dx = -e^x \cos x + \int e^x \cos x \, dx$

NOTE* I can reuse u and v because we started a new problem. However, if you are linking questions, you may need to use s, t instead of u₁, v₁ and u₂, v₂. So that you do not have u and v equaling two different things. $v_1 \ne v_2$ .

$-e^x \cos x + \int e^x \cos x \, dx$

$u_2 = e^x$ $du_2 = e^x dx$

$dv_2 = \cos x \, dx$ $v_2 = \sin x$

$-e^x \cos x + \int e^x \cos x \, dx = -e^x \cos x + \left(e^x \sin x - \int e^x \sin x \, dx \right)$

*At this point, many students give up but this is the answer with one more move!

$\int e^x \sin x \, dx = -e^x \cos x + e^x \sin x - \int e^x \sin x \, dx$

$\Rightarrow I = -e^x \cos x + e^x \sin x - I$

$\Rightarrow \int e^x \sin x \, dx + \int e^x \sin x \, dx = -e^x \cos x + e^x \sin x$

$\Rightarrow 2 \int e^x \sin x \, dx = -e^x \cos x + e^x \sin x$

$\therefore \int e^x \sin x \, dx = \frac{1}{2} \left[-e^x \cos x + e^x \sin x \right] + C$

$\Rightarrow \int e^x \sin x \, dx = \frac{e^x}{2} [\sin x - \cos x] + C$

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TEST PREP QUESTION Case 5: Exponential and Trigonometric

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