I am trying to solve the question $$ \int\sqrt{\sin x+\cos x}\,dx $$

Is their any substitution by which I can get the answer. I tried different substitution like i multiplied both numerator and denominator by $\sqrt{\sin x+\cos x} $ and uses sinx -cos x=t but getting complicated

  • 4
    $\begingroup$ Elliptic integral! $\endgroup$
    – Mikasa
    Sep 10, 2018 at 7:40
  • 6
    $\begingroup$ To clarify mrs' comment (if that is necessary), a basic trigonometric identity is $\sin x + \cos x = \sqrt{2} \sin\left( x+\frac{\pi}{4}\right)$, and the integral of $\sqrt{\sin x}$ is a well-known elliptic integral which has no closed form in terms of elementary functions. $\endgroup$
    – Al Nejati
    Sep 10, 2018 at 8:06
  • $\begingroup$ Does your original question specify bounds? $\endgroup$
    – Deepak
    Sep 10, 2018 at 11:44

2 Answers 2


Elliptic Integral of the Second Kind:

$$E(\varphi,k) = \int_0^\varphi \sqrt {1 - k^2(\sin\vartheta)^2}\, \mathrm d\vartheta$$

$$ I = \int \sqrt{\sin(x) + \cos(x)}\,dx$$ $$\sqrt[4]{2}\int \sqrt{\frac{1}{\sqrt{2}}\sin(x) + \frac{1}{\sqrt{2}}\cos(x)}\,dx$$ $$\sqrt[4]{2}\int \sqrt{\sin(x)\cos\left(\frac{\pi}{4}\right) + \cos(x)\sin\left(\frac{\pi}{4}\right)}\,dx$$ Set $u = \left(x + \frac{\pi}{4}\right) \to du = dx$ $$\sqrt[4]{2}\int \sqrt{\sin(u)}\,du$$ So the elliptic integral can also be written as $$\int \sqrt{\sin(u)}\,du = \int \sqrt{1 - 2\sin^2\left(\frac{\pi}{4}-\frac{\pi}{2}\right)}\,du$$ $$-2\int\sqrt{1-2\sin^2(u)}$$ $$-2\,E(u\,|\,2) + C$$ $$-2\,E\left(\frac{\pi}{4} - \frac{u}{2}\,|\,2\right) + C$$ Back substitution $u = \left(x + \frac{\pi}{4}\right)$ $$-2\,E\left(\frac{\pi}{4} - \frac{\left(x + \frac{\pi}{4}\right)}{2}\,|\,2\right) + C$$ $$-2\,E\left(\frac{\pi}{8} - \frac{x}{2}\,|\,2\right) + C$$ $$-2\,E\left(\frac{1}{8}\left(\pi - 4x\right)\,|\,2\right) + C$$ Therefore

$$I = -2\sqrt[4]{2}\,E\left(\frac{1}{8}(\pi - 4x)\,|\,2\right) + C$$


$$I=\int\sqrt{\sin(x)+\cos(x)}dx$$ let: $$\sin(x)+\cos(x)=R\sin(x+a)$$ $$\sin(x)+\cos(x)=R\sin(x)\cos(a)+R\cos(x)\sin(a)$$ so: $R\cos(a)=1$ and $R\sin(a)=1 \, \therefore \tan(a)=1$ and $R^2=2$

we then get: $R=\sqrt{2}$ and $a=\frac{\pi}{4}$

the integral can now be rewritten as: $$I=\int\sqrt{\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)}dx=2^\frac{1}{4}\int\sqrt{\sin\left(x+\frac{\pi}{4}\right)}dx$$ let $u=x+\frac{\pi}{4} \,\therefore\,du=dx$ $$I=2^\frac{1}{4}\int\sqrt{\sin(u)}du$$ then this has no elementary solution but can be defined using the incomplete elliptic integral of the second kind.


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