Solutions - 0010-06

(1) Select $u$ and $v^{\prime }$ considering LIATE:

$$\begin{array}{cc}u=\sin x& v^{\prime }=e^x\\ u^{\prime }=\cos x& v=e^x\end{array}\begin{array}{c}⟶\int u\cdot v^{\prime }=\int \sin x\cdot e^x\text{original}\\ ⟶\int u^{\prime }\cdot v=\int \cos x\cdot e^x\text{final}\end{array}$$

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(2) Apply IBP formula $uv-\int u^{\prime }\cdot v$ and compute integral:

$$uv-\int u^{\prime }\cdot v\gg \gg \sin x\cdot e^x-\int \cos x\cdot e^{x}dx$$

Therefore:

$$\begin{array}{ccc}& \int \sin x\cdot e^{x}dx=\sin x\cdot e^x-\int \cos x\cdot e^{x}dx& \text{(Eqn. A)}\end{array}$$

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(3) Repeat. Select $u$ and $v^{\prime }$ considering LIATE:

$$\begin{array}{cc}u=\cos x& v^{\prime }=e^x\\ u^{\prime }=-\sin x& v=e^x\end{array}\begin{array}{c}⟶\int u\cdot v^{\prime }=\int \cos x\cdot e^x\text{original}\\ ⟶\int u^{\prime }\cdot v=\int -\sin x\cdot e^x\text{final}\end{array}$$

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(4) Apply IBP formula $uv-\int u^{\prime }\cdot v$ and compute integral:

$$\begin{array}{cc}uv-\int u^{\prime }\cdot v& \gg \gg \cos x\cdot e^x-\int -\sin x\cdot e^{x}dx\\ & \gg \gg \cos x\cdot e^x+\int \sin x\cdot e^{x}dx\end{array}$$

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(5) Now insert in Eqn. A:

$$\int \sin x\cdot e^{x}dx=\sin x\cdot e^x-(\cos x\cdot e^x+\int \sin x\cdot e^{x}dx)$$

Introduce notational label:

$$I=\int \sin x\cdot e^{x}dx$$

Now use this label in Eqn. A and solve for $I$:

$$\begin{array}{c}(A)\gg \gg I=(\sin x-\cos x)e^x-I\\ \\ \gg \gg 2I=(\sin x-\cos x)e^x\\ \\ \gg \gg I=\frac{1}{2}(\sin x-\cos x)e^x+C\end{array}$$