Hi, you might have come across:$\displaystyle \int e^{ax}\cos bx$ dx,

How do you solve it? You might use integration by parts, and also complex numbers, and I find use of complex numbers interesting!

Say $A = \displaystyle \int e^{ax}\cos bx$dx,

and $B = \displaystyle \int e^{ax}\sin bx$ dx

Hence, $A + iB = \displaystyle \int e^{ax} (\cos bx + i \sin bx)$dx = $\displaystyle \int e^{ax} (e^{i bx})$dx

= $\displaystyle \int e^{(a+ib)x}$dx

= $\displaystyle \frac{e^{(a+ib)x}}{a+ib}$

= $\displaystyle \frac{e^{(a+ib)x}(a - ib)}{a^2 + b^2}$

$\Rightarrow A + iB = z = \displaystyle \frac{e^{ax} (\cos bx + i \sin bx)(a - ib)}{a^2+b^2}$

Clearly,

$A = \text{Re}(z) = \displaystyle \frac{e^{ax}}{a^2+b^2} (a \cos bx + b \sin bx)$

$B = \text{Im}(z) = \displaystyle \frac{e^{ax}}{a^2+b^2} (a \sin bx - b \cos bx)$

You can try to find this definite integral:

**Problem:** $\displaystyle \int_{0}^{\pi} e^{(\cos x)} \cos(\sin x)$ dx

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## Comments

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TopNewestShort and Sweet! :)

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Thanks!

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$cos { (\sin { x) } } =\quad ({ e }^{ isinx }+{ e }^{ -isinx })/2\\ { e }^{ cosx }\cos { (\sin { x) } } =\quad ({ e }^{ cosx+isinx }+{ e }^{ cosx-isinx })/2=({ e }^{ { e }^{ ix } }+{ e }^{ { e }^{ -ix } })/2\\ =((1+\frac { { e }^{ ix } }{ 1! } +\frac { { e }^{ i2x } }{ 2! } +...)+(1+\frac { { e }^{ -ix } }{ 1! } +\frac { { e }^{ -2ix } }{ 2! } +...))/2$

In 2 more steps you will get the answer. The answer in this case is $pi$

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Solve his other question too: Problem without words !

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Hi Jatin! Thanks for the great post, but just one thing I was wondering that incorporating i, the complex number, and using the usual laws of Calculus, is it mathematically correct, I mean the laws of calculus is for reals. I may be fundamentally wrong somewhere but I need the answer.

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Awesome article.

By using your method I arrive at a step from where I can't proceed further.

The step is: $A+iB=\int (e)^{e^{ix}}\,dx$, where $A=\int (e)^{(cos x)}cos(sin x)\,dx$ and $B=\int (e)^{(cos x)}sin(sin x)\,dx$.

How to do after this?Precisely,what's the real part?

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Hi, Bhargav, you are close, try to use expansion for $e^x$.

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Is the answer $\pi$?

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$B$?

I reach the same answer but is it possible to evaluateLog in to reply

You came our with a simple and an interesting solution the technique that triggered me was to solve it using By Parts Method but have to admit your approach was far far better than mines Can you tell me how do you get such ideas at and tender age of 15(Just Asking)?

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Hi, solving this integral using complex numbers is well(not very much well though) known. I did not come up with it myself.

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Yeah, actually the same technique's been discussed with us at our insti as well... anyways, it's really good...

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Awesome!!

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-e^{-1} -1 is the ans

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great..:-)

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Too beautiful

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Answer is π

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Thanks a ton..!

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Easy approach....nice trick...

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Interesting method!

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Finally I got it! :)

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THANKS. VERY POWERFULL METHOD-CAN BE TACKLED ANY COMPLICATED INTEGRALS LIKE THIS ONE.

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It is an amazing approach to solve such problems.

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$W0w$ $£xcellent approach$

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