Integrating $\int e^{\cos x} \cos(2x+\sin x)dx$

Question

What is the result of this integral? $$\int e^{\cos x} \cos(2x+\sin x)dx$$

I tried to use integration by parts but nothing made sense to me in this given problem. What should we do to solve these kinds of integrals? This is from a high school contest so we cannot use complex integration. Thank you!

Answer 1

$\int e^{\cos x}\cos(2x+\sin x)dx$

$=\int e^{\cos x}\cos(x+(x+\sin x))dx$

$=\int e^{\cos x}(\cos(x)\cos(x+\sin x)-\sin x \sin(x+\sin x))dx$

$=\int e^{\cos x}(-\sin x)\sin(x+\sin x)dx+\int e^{\cos x}(\cos x)\cos(x+\sin x)dx$

$=e^{\cos x}\sin(x+\sin x)-\int e^{\cos x}\cos(x+\sin x)(1+\cos x)dx+\int e^{\cos x}(\cos x)\cos(x+\sin x)dx$

$=e^{\cos x}\sin(x+\sin x)-\int e^{\cos x}\cos(x+\sin x)dx$

$=e^{\cos x}\sin(x+\sin x)-\int e^{\cos x}(\cos x\cos(\sin x)-\sin x\sin(\sin x))dx$

$=e^{\cos x}\sin(x+\sin x)-\int e^{\cos x}(-\sin x)\sin(\sin x)dx-\int e^{\cos x}(\cos x)\cos(\sin x)$

$=e^{\cos x}\sin(x+\sin x)- e^{\cos x}\sin(\sin x)+\int e^{\cos x}(\cos x)\cos(\sin x)dx-\int e^{\cos x}(\cos x)\cos(\sin x)dx$

$=e^{\cos x}\left(\sin(x+\sin x)- \sin(\sin x)\right)+c$

Answer 2

\begin{align} &\int e^{\cos x}\cos(2x+\sin x) dx\\ =& \ \frac{d}{da} \bigg(\int e^{a\cos x} \cos(x+a\sin x) dx \bigg)\bigg|_{a=1} \\ =&\ \frac{d}{da}\bigg( \frac1a e^{a\cos x}\sin(a\sin x) \bigg)\bigg|_{a=1} =e^{\cos x}[\sin(x+\sin x)-\sin(\sin x)] \end{align}

Answer 3

Noting that

$$\int e^{\cos x} \cos (2 x+\sin x) d x = \Re \int e^{\cos x} e^{i(2 x+\sin x)} d x$$ We first evaluate $$ \begin{aligned} \int e^{\cos x} e^{i(2 x+\sin x)} d x & =\int e^{\cos x+2 x i+i \sin x} d x \\ & =\int e^{e^{x i}} \cdot e^{2 x i} d x \\ & =-i \int e^{x i} d\left(e^{e^{x i}}\right) \\ & =-i e^{x i} e^{e^{x i}}+i \int i e^{x i} e^{e^{x i}} d x \\ & =-i e^{x i} e^{e^{x i}}+i e^{e^{x i}}+C \\ & =-i e^{e^{x i}}\left(e^{x i}-1\right)+C \end{aligned} $$

$$ \begin{aligned} \int e^{\cos x} \cos (2 x+\sin x) d x =& \Im\left[e^{e^{x i}}\left(e^{x i}-1\right)\right]+C\\=& \Im\left[e^{\cos x+i \sin x} \cdot(\cos x+i \sin x-1)\right]+C\\=& \Im\left[e^{\cos x}(\cos(\sin x)+i \sin (\sin x))(\cos x+i \sin x-1)\right]+C\\=& e^{\cos x}[\sin x \cos (\sin x)+\sin (\sin x)(\cos x-1)]+C\\=& 2 e^{\cos x} \sin \frac{x}{2}\left[\cos (\sin x) \cos \frac{x}{2}-\sin (\sin x) \sin \frac{x}{2}\right]+C\\=& 2 e^{\cos x} \sin \frac{x}{2} \cos \left(\frac{x}{2}+\sin x\right)+C \end{aligned} $$ As a bonus, we use the imaginary parts and get $$ \int e^{\cos x} \sin (2 x+\sin x)dx =\Im \int e^{\cos x} e^{i(2 x+\sin x)} d x =2 e^{\cos x} \sin\frac{x}{2}\sin \left(\frac{x}{2}+\sin x\right)+C $$

Answer 4

I think we can improvise upon the question:

If $I_n=\int e^{\cos x}\cos(nx+\sin x)dx$

then

$I_n+(n-1)I_{n-1}=$

(A) $e^{\cos x}\sin(x+\sin x)+c$

(B) $e^{\cos x}\sin((n-1)x+\sin x)+c$

(C) $e^{\cos x}\sin((n-1)x+ n\sin x)+c$

(D) $e^{\cos x}\sin(nx+(n-1)\sin x)+c$

Answer 5

For the fun (since you already received good answers), expand the cosine and use the tangent half-angle substitution to end with $$I=2\int \frac{ 4 t \left(t^2-1\right) \sin \left(\frac{2 t}{t^2+1}\right)+\left(t^4-6 t^2+1\right) \cos \left(\frac{2 t}{t^2+1}\right)}{\left(t^2+1\right)^3} \,e^{\frac{1-t^2}{1+t^2}}\,dt$$ and, if you are very patient $$I=\frac{2 t}{t^2+1}\left(\cos \left(\frac{2 t}{t^2+1}\right)-t \sin \left(\frac{2t}{t^2+1}\right)\right)\,e^{\frac{1-t^2}{1+t^2}}$$

Now, back to $x$ for the result.