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The integral is :$$I=\int e^{\alpha x}\cos\beta xdx$$

To evaluate the integral I used integral by parts method twice. for first integral I used substitution $u=e^{\alpha x}$ and $dv=\cos\beta x dx$ and for second one $dv=\sin\beta x$ My final answer is $$I=\frac{\beta}{\alpha^2+\beta^2}e^{\alpha x}(\sin\beta x+\frac{\alpha}{\beta}\cos\beta x)+C$$

But the answer in the book I am studying is :

$$I=\frac{\alpha}{\alpha^2+\beta^2}e^{\alpha x}(\cos\beta x+\frac{\beta}{\alpha}\sin\beta x)+C$$

First I thought my answer is wrong but I took derivative of that and obtained $e^{\alpha x}\cos\beta x$. so I realized my answer is also correct.

I find out the reason for different answers is different substitution the book used $u=\cos\beta x$ and $dv=e^{\alpha x}dx$. My question is why we got different answers here for different $u$ and $dv$? Is there any mathematically logic that can explain this?

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    $\begingroup$ Move $\beta$ inside the parantheses for the first version and $\alpha$ inside for the second. $\endgroup$ Jan 10, 2021 at 7:04
  • $\begingroup$ @Semiclassical Oh they are the same. never thought about that! $\endgroup$
    – Etemon
    Jan 10, 2021 at 7:05

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If $F'(x)=G'(x)$ then we have $F(x)-G(x)=constant$ so the difference related to constant. For example since we have $$\sin^2x+\cos^2x=1$$ then one can let $\sin^2x=1-\cos^2x$ or $\cos^2x=1-\sin^2x$ that is no basic differnce between $\sin^2x $ and $\cos^2x$ as answer of integrals.

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    $\begingroup$ I realized they are the same expression as mentioned by Semiclassical in the comments. So it is not the case. $\endgroup$
    – Etemon
    Jan 10, 2021 at 10:35
  • $\begingroup$ Also there is different between $\sin^2 x$ and $\cos^2 x$ as answer of integral. because one is $\sin^2 x$ plus a constant and another is $\color{red}{-}\sin^2 x$ plus a constant. $\endgroup$
    – Etemon
    Jan 10, 2021 at 10:47

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