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Suppose we begin to integrate $e^xcosh(x)\ dx$ by parts. Regardless of our $u$ and $dv$ choices, we arrive at $$\int e^x\cosh x\ dx = e^x\cosh(x) - e^x\sinh(x) + \int e^x\cosh x\ dx\tag{Equation 1}$$ If we subtract the integral from both sides of $(\text{Equation }1)$, leaving behind only an integration constant $C$, we get $$C = e^x\cosh(x) - e^x\sinh(x)\tag{Equation 2}$$ Graphing $e^x\cosh(x) - e^x\sinh(x)$ would reveal that $C = 1$, but what's interesting is that even without graphing, or applying any identities, we can learn from the left side of $(\text{Equation }2)$ that the value of $e^x\cosh(x) - e^x\sinh(x)$ is a constant; it doesn't depend on $x$. It's intriguing that such an unusual fact would emerge, while it had nothing to do with the original problem.

Are there any other instances where canceling out the original integral during integration by parts is profitable, either for solving the integral, or discovering some external fact?

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    $\begingroup$ Try to integrate $\int e^x\cos x dx$ or $\int e^x\sin x dx$ by parts. The integral does not cancel but it is useful to solve for the integral. $\endgroup$ – zxcvber Jul 2 '18 at 5:37
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Lots of integrals involving trigonometric functions will have this property. Consider the integral $$ \int f(x) g'(x) \,\,\mathrm{d}x.\tag{1} $$ Using integration by parts, this integral becomes $$ f(x)g(x) - \int f'(x)g(x)\,\,\mathrm{d}x.\tag{2} $$ Now, suppose that $f$ is a function such that $f'(x) = k_{1}\cdot f(x)$, and similarly $g'(x) = k_{2} \cdot g(x)$. One such example of a function is $f = \exp(k_{1}x)$. Then combining $(1)$ and $(2)$ under these conditions gives $$ \int f(x)g'(x) \,\,\mathrm{d}x = \frac{f(x)g'(x)}{k_{2}} - \frac{k_{1}}{k_{2}} \int f(x)g'(x) \,\,\mathrm{d}x, $$ so that $$ \Big(1 + \frac{k_{1}}{k_{2}}\Big) \int f(x)g'(x) \,\,\mathrm{d}x = \frac{f(x)g'(x)}{k_{2}} \implies \int f(x)g'(x) \,\,\mathrm{d}x = \frac{f(x)g'(x)}{k_{1} + k_{2}} \color{red}{+ c}. $$ This makes sense since only the exponential function is equal to its derivative, so $f$ and $g$ must both be exponential functions and thus $fg'$ is exponential with $f(x)g'(x) = \exp\big((k_{1} + k_{2})x\big)$ up to a scalar multiple.


A more interesting question is when $f(x)$ is a scalar multiple of $f'(x)$, but $g(x)$ is a scalar multiple of $g''(x)$. Examples for $g$ include both $\sin(x)$ and $\cos(x)$ (and also $e^{x}$). So, suppose that $f'(x) = k_{1}\cdot f(x)$ and that $g''(x) = k_{2}^{2}\cdot g(x)$. Consider the integral $$ \int f(x)g''(x) \,\,\mathrm{d}x.\tag{3} $$ This integral becomes $$ f(x)g'(x) - \int f'(x)g'(x) \,\,\mathrm{d}x = f(x)g'(x) - \Big( f'(x)g(x) - \int f''(x)g(x) \,\,\mathrm{d}x \Big).\tag{4} $$ Combining $(3)$ and $(4)$ gives $$ \int f(x)g''(x) \,\,\mathrm{d}x = f(x)g'(x) - f'(x)g(x) + \int f''(x)g(x) \,\,\mathrm{d}x.\tag{5} $$ Notice that this is a very similar form to your question. Now applying the derivative equalities we see that $(5)$ becomes $$ \int f(x)g''(x) \,\,\mathrm{d}x = f(x)g'(x) - f'(x)g(x) + \frac{k_{1}^{2}}{k_{2}^{2}} \int f(x)g''(x) \,\,\mathrm{d}x, $$ so that $$ \left( 1 - \Big(\frac{k_{1}}{k_{2}}\Big)^{2}\right)\int f(x)g''(x) \,\,\mathrm{d}x = f(x)g'(x) - f'(x)g(x), $$ and thus $$ \int f(x)g''(x) \,\,\mathrm{d}x = \frac{f(x)g'(x) - f'(x)g(x)}{1 - \Big(\frac{k_{1}}{k_{2}}\Big)^{2}} \color{red}{+ c}. $$ This method/formula is particularly helpful precisely when $f$ and $g$ are periodic in their derivatives, as one can easily get caught up doing repeated integration by parts without getting anywhere (at least, I know that I have). This method/formula was particularly helpful for me in evaluating integrals like $$ \int e^{-st}\cos(3t) \,\,\mathrm{d}t, $$ which is the Laplace transform of $\cos(3t)$, so that this method is helpful for finding solutions to differential equations.

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$2\cosh x=e^x+e^{-x}$ and $2\sinh x=e^x-e^{-x}$, hence $e^x\cosh(x) - e^x\sinh(x)=1$.

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    $\begingroup$ This does not answer the question... $\endgroup$ – user193810 Jul 2 '18 at 11:49

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