I was trying to find out the total kinetic energy of a sinusoidal wave - $y(x,t) =A\sin(wt-kx)$.
The expression I got is - $KE_{tot} =\displaystyle\int_{_{A\sin(wt-kx)} }\mu Aw\cos(wt-kx)\mathrm{ds}$
which simplifies to - $\displaystyle\int{\mu Aw\cos(wt-kx)\sqrt{1+(Ak\cos(wt-kx))^2}}\mathrm{dx}$
I' stuck here with this integral : $\displaystyle\int{\cos x\sqrt{1+a^2\cos^2x}}\mathrm{dx}$ a being a constant. Please help.
If $a=0$, this problem is easily solved. Assuming $a\neq 0$, we can make the substitution $\cos^2(x)=1-\sin^2(x)$, so the integral becomes: $$ \int \cos(x)\sqrt{-a^2\sin^2(x)+a^2+1}dx$$ Let $u=\sin(x)\implies \frac{du}{dx}=\cos(x)\implies dx = \frac{du}{cos(x)}$, we arrive at $$ \int\sqrt{-a^2u^2+a^2+1}du$$ Now, we recognize we can use a trig substitution: $u=\frac{\sqrt{a^2+1}\sin(v)}{a}\implies du = \frac{\sqrt{a^2+1}\cos(v)dv}{a}$, and we also have that $v =\arcsin(\frac{au}{\sqrt{a^2+1}})$ so our integral becomes: $$ \int \frac{\sqrt{a^2+1}\cos(v)\sqrt{-(a^2+1)\sin^2(v)+a^2+1}}{a}dv$$ Simplifying using $-(a^2+1)\sin^2(v)+a^2+1= (a^2+1)\cos^2(v)$, we arrive at: $$\frac{a^2+1}{a}\int \cos^2(v)dv = \frac{a^2+1}{a}\left(\frac{\cos(v)\sin(v)}{2}+\frac{v}{2}\right) $$ Substituting back in $v =\arcsin(\frac{au}{\sqrt{a^2+1}})$, and using the fact that $\sin(\arcsin(\frac{au}{\sqrt{a^2+1}}))=\frac{au}{\sqrt{a^2+1}}$, $\cos(\arcsin(\frac{au}{\sqrt{a^2+1}}))=\sqrt{1-\frac{a^2u^2}{a^2+1}}$ and $u = \sin(x)$, we arrive at the final answer: $$ \int \cos(x)\sqrt{a^2\cos^2(x)+1}dx = \frac{(a^2+1)\arcsin(\frac{a\sin(x)}{\sqrt{a^2+1}})}{2a}+\frac{\sqrt{a^2+1}\sin(x)\sqrt{1-\frac{a^2\sin^2(x)}{a^2+1}}}{2}+ C, \quad C\in \mathbb{R}$$