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How would I evaluate the integral: $$I=\int{(\cos(x)\cosh(x)+\sin(x)\sinh(x)})\,dx$$ My thought was to use: $$\cos(ix)=\cosh(x)$$ and $$\sin(ix)=i\sinh(x)$$ or expand all four trig functions into exponentials but this was very messy

EDIT:

If I split this into two integrals where $I=I_1+I_2$ $$I_1=\int{cos(x)cosh(x)}dx$$ $$I_2=\int{sin(x)sinh(x)}dx$$

$$I_1=sin(x)cosh(x)-\int{sin(x)sinh(x)}dx$$ This second part is equal to $I_2$ so does that mean that $$I=sin(x)cosh(x)$$

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    $\begingroup$ Have you tried integration by parts? $\endgroup$
    – Dylan
    Commented May 23, 2018 at 19:09
  • $\begingroup$ as in split into two integrals and go from there? $\endgroup$
    – Henry Lee
    Commented May 23, 2018 at 19:11
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    $\begingroup$ Integrate by parts one of the terms. It'll become really obvious... $\endgroup$
    – Dylan
    Commented May 23, 2018 at 19:13
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    $\begingroup$ Expanding the integrand as exponentials isn't that messy. I tried it and got $\frac{1}{4} \left[ (1+i) (e^{x-ix} + e^{-x+ix}) + (1-i) (e^{x+ix} + e^{-x-ix}) \right]$. $\endgroup$ Commented May 23, 2018 at 19:15
  • $\begingroup$ While grossly inefficient here, my first move would have been to try the same as the removed comment suggested: write the integrand $\cos x\cos ix+i\sin x\sin ix$ and linearize. $\endgroup$ Commented May 23, 2018 at 19:17

4 Answers 4

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Hint...try differentiating $\sin x\cosh x$

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    $\begingroup$ (+1) for the monster hint. $\endgroup$
    – Mark Viola
    Commented May 23, 2018 at 19:14
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    $\begingroup$ Not a hint as much as just giving them the answer $\endgroup$
    – Dylan
    Commented May 23, 2018 at 19:15
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    $\begingroup$ I'm actually going to downvote this. $\endgroup$
    – Dylan
    Commented May 23, 2018 at 19:20
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    $\begingroup$ @YvesDaoust It would be better to tell them how to make the ansatz, rather than give them the ansatz straight away. How can you apply this knowledge without knowing all the steps? $\endgroup$
    – Dylan
    Commented May 23, 2018 at 19:47
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    $\begingroup$ @DavidQuinn To me it's like spoiling a movie. Once you gave this hint, it's obvious that it's going to be the answer. In the end he still wouldn't know how to go back and do the integral. All he knows is the answer to the problem. Anyway, he followed my hint of IBP and came to the anti-derivative, so that answers your question (of whether he knows the answer now). $\endgroup$
    – Dylan
    Commented May 23, 2018 at 19:50
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I think you might want to notice that we have $\sin x$ on right side while it's derivative i.e. $\cos x$ on left side. Also we have $\cosh x$ on left side while it's derivative i.e. $\sinh x$ on the right side.

Don't you think this seems to be much similar to what happens in the product rule.

Let $f(x)=\sin x$ and $g(x)=\cosh x$

Hence the integral can be written as $$\int (f'g +g'f) dx$$

Which simply equals $$f(x)\cdot g(x)+ C$$ by noticing the product rule.

Hence the answer to integral is $$\sin x\cosh x +C$$

You might also want to solve One such question from MIT Integration bee using similar idea which is $$\int (\sin (101x) \cdot \sin^{99}x)dx$$

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    $\begingroup$ Off-Topic: I found that MIT Integration Bee Integral!! youtu.be/qQ-56b_LvOw at 1:17:35. That was amazing! $\endgroup$
    – manooooh
    Commented May 24, 2018 at 2:18
  • $\begingroup$ $\int{sin(101x).sin^{99}(x)}dx=\int{sin^{99}(x).cos(x).sin(100x)+sin^{100}(x).cos(100x)}dx$ and $\int{sin^{99}(x).cos(x)}dx=\frac{sin^{100}(x)}{100}$ then use the same method as was used before where the second part of the integral $I_2$ is equal to the product of doing IBP on $I_1$ $\endgroup$
    – Henry Lee
    Commented May 24, 2018 at 8:03
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    $\begingroup$ @HenryLee Or else take $f(x)=\sin^{100} x$ and $g(x)=\sin (100x)$ and continue $\endgroup$ Commented May 24, 2018 at 8:10
  • $\begingroup$ it is quite a nice integral though, thats why I often do the MIT integration bee integrals, and the original one was from the qualifying test: mit.edu/~same/pdf/qualifying_round_2018_test.pdf $\endgroup$
    – Henry Lee
    Commented May 24, 2018 at 8:13
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Your integrand is not exactly in a suitable form. You'd better consider the following two functions:

$$\cos x\cosh x-i\sin x\sinh x=\cos(x+ix)=\cos(1+i)x,$$ and $$\cos x\cosh x+i\sin x\sinh x=\cos(x-ix)=\cos(1-i)x.$$

The antiderivatives are simply

$$\frac{\sin(1+i)x}{1+i},\frac{\sin(1-i)x}{1-i}$$ which you can develop in terms of the trigonometric and hyperbolic functions. Finally, take the sum of the real and imaginary parts. And laugh.

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$$\int \cos(x) \cosh(x)+\sin(x) \sinh(x)dx$$ First divide this into two integrals $$\int \cos(x) \cosh(x)dx+\int \sin(x) \sinh(x) dx$$ Then solve for $\int \cos(x) \cosh(x)dx$ using integration by parts and we get, $$\int \cos(x) \cosh(x)dx=\frac12( \cos(x) \sinh(x)+\sin(x)\cosh(x))$$ and $$\int \sin(x) \sinh(x)dx=\frac12(\sin(x)\cosh(x)-\cos(x)\sinh(x))$$

Now add them both together and we get, $$\int \cos(x) \cosh(x)+\sin(x) \sinh(x)dx=\frac12( \cos(x) \sinh(x)+\sin(x)\cosh(x))+\frac12(\sin(x)\cosh(x)-\cos(x)\sinh(x))$$ $$=\frac12( \cos(x) \sinh(x)+\sin(x)\cosh(x)+\sin(x)\cosh(x)-\cos(x)\sinh(x))$$ $$=\frac12(2 \sin(x)cosh(x))$$ $$\int \cos(x) \cosh(x)+\sin(x) \sinh(x)dx=\sin(x)\cosh(x)+C$$

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