Cool Integration Techniques! (Part 1)

Calculus Level 4

0 2 π e cos θ cos ( sin θ ) d θ = ? \large \int_{0}^{2\pi}{{e}^{\cos\theta} \cos(\sin\theta) \, d\theta} = \, ?


The answer is 6.28.

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3 solutions

Aniket Sanghi
Oct 31, 2016

Let F ( x ) = e c o s x . c o s ( s i n x ) = R . P . O e c o s x . e i s i n x = R . P . O e c o s x + i s i n x = R . P . O e e i x = R . P . O 1 + e i x + e 2 i x 2 ! + e 3 i x 3 ! + . . . . . . . F (x) = {e}^{cosx} . cos(sinx) \\ \qquad = R.P.O \quad {e}^{cosx} . {e}^{isinx} \\ \qquad = R.P.O \quad {e}^{cosx + isinx} \\ \qquad = R.P.O \quad {e}^{{e}^{ix}}\\ \qquad = R.P.O \quad 1 + {e}^{ix} + \frac{ {e}^{2ix}}{2!} + \frac{{e}^{3ix}}{3!} + .......\\

R . P . O = = R.P.O == Real Part of

Hence you will get

0 2 π F ( x ) = 0 2 π 1 + c o s x + c o s 2 x 2 ! + c o s 3 x 3 ! + . . . . . = 2 π + 0 + 0 + . . . . = 2 π \int\limits_{0}^{2\pi}{F (x)} = \int\limits_{0}^{2\pi}{1 + cosx + \frac{cos2x}{2!} + \frac{cos3x}{3!} + .....} = 2\pi + 0 + 0 + .... = 2\pi

c o s ( s i n x ) = e i s i n x + e i s i n x / 2 cos(sinx)=e^{isinx}+e^{-isinx}/2

I cant understand second line

Kushal Bose - 4 years, 7 months ago

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e i s i n x = c o s ( s i n x ) + i s i n ( s i n x ) S o , R e a l p a r t o f e i s i n x = c o s ( s i n x ) {e}^{isinx} = cos (sinx) + i sin (sinx) \\ So, Real \quad part \quad of \quad {e}^{isinx} = cos (sinx)

Aniket Sanghi - 4 years, 7 months ago

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yess clear

Kushal Bose - 4 years, 7 months ago

Nice method! +1

Harsh Shrivastava - 4 years, 7 months ago

The real part of the integral integral can be written as I = 0 2 π e e i θ d θ \displaystyle I=\int_{0}^{2\pi}e^{e^{i\theta}} d\theta

Now putting e i θ = z e^{i\theta}=z and integrating around the unit circle we get,

I = z = 1 i e z z d z = 2 π i × ( i ) = 2 π \displaystyle I=\int_{|z|=1} \frac{-ie^z}{z}dz=2\pi i\times(-i)=\boxed{2\pi}

Alternate solution could be resorting to Feynman's way of Integration by considering the function F ( x ) = 0 2 π e x cos θ cos ( x sin θ ) d θ F(x)=\large \int_{0}^{2\pi}{{e}^{x\cos\theta} \cos(x\sin\theta) \, d\theta} and then proceeding.

Our required integrand is F ( 1 ) F(1) .

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