2
$\begingroup$

Solving the integral : $$\int_0^{\frac{\pi}{2}}\frac{\sin x}{1+\sqrt{\sin 2x}}dx=?$$


My try:

$$\int_0^{\frac{\pi}{2}}\frac{\sin x}{1+\sqrt{\sin 2x}}\cdot \frac{1-\sqrt{\sin 2x}}{1-\sqrt{\sin 2x}}dx$$

$$\int_0^{\frac{\pi}{2}}\frac{\sin x(1-\sqrt{\sin 2x})}{1-\sin 2x}dx$$

$$(\sin x-\cos x)^2=1-\sin 2x$$

So :

$$\int_0^{\frac{\pi}{2}}\frac{\sin x(1-\sqrt{\sin 2x})}{(\sin x-\cos x)^2}dx$$

Now what?

$$$$

$\endgroup$
5
$\begingroup$

The given integral equals $$ \frac{1}{2}\int_{0}^{\pi}\frac{\sin\tfrac{x}{2}}{1+\sqrt{\sin x}}\,dx =\frac{1}{2\sqrt{2}}\int_{0}^{\pi}\frac{\sqrt{1-\cos x}}{1+\sqrt{\sin x}}\\=\frac{1}{2\sqrt{2}}\int_{0}^{\pi/2}\frac{\sqrt{1-\cos x}+\sqrt{1+\cos x}}{1+\sqrt{\sin x}}\,dx\\ =\frac{1}{2}\int_{0}^{\pi/2}\frac{\sqrt{1+\sin x}}{1+\sqrt{\sin x}}\,dx$$ or $$ \frac{1}{2}\int_{0}^{1}\frac{\sqrt{1+x}}{\sqrt{1-x^2}(1+\sqrt{x})}\,dx = \frac{1}{2}\int_{0}^{1}\frac{dx}{(1+\sqrt{x})\sqrt{1-x}}=\int_{0}^{1}\frac{x\,dx}{(1+x)\sqrt{1-x^2}}$$ which equals $\frac{\pi}{2}-\int_{0}^{1}\frac{dx}{(1+x)\sqrt{1-x^2}}$. On the other hand $$ \int_{0}^{1}\frac{dx}{(1+x)\sqrt{1-x^2}}=\int_{0}^{\pi/2}\frac{d\theta}{1+\sin\theta}=\int_{0}^{\pi/2}\frac{d\theta}{1+\cos\theta}=\int_{0}^{\pi/2}\frac{d\theta}{2\cos^2\tfrac{\theta}{2}}=1, $$ hence: $$\boxed{\int_{0}^{\pi/2}\frac{\sin x}{1+\sqrt{\sin(2x)}}\,dx = \color{blue}{\frac{\pi}{2}-1},} $$ no particular elliptic integral like $K\left(\frac{1}{2}\right)$ or $E\left(\frac{1}{2}\right)$ is really involved.

$\endgroup$
2
$\begingroup$

$$I=\int_0^{\frac{\pi}{2}}\frac{\sin x}{1+\sqrt{\sin 2x}}\,dx$$ Using the fact that: $$\int_a^b f(x) dx=\frac{1}{2} \int_a^b (f(x)+f(a+b-x))dx$$ $$\Rightarrow I=\frac12 \int_0^\frac{\pi}{2} \frac{\sin x+\cos x}{1+\sqrt{\sin 2x}}dx$$Now let's substitute $\,\sin x-\cos x=u\,$ $\rightarrow u^2=1-\sin(2x)$$\rightarrow\sin(2x)=1-u^2$ $$I=\int_{0}^{1}\frac{du}{1+\sqrt{1-u^2}}$$ Letting $u=\sin t$ yields: $$I=\int_0^{\frac{\pi}{2}}\frac{\cos t}{1+\cos t}\,dt= \int_0^{\frac{\pi}{2}}\left(\frac{1+\cos t}{1+\cos t}-\frac{1}{1+\cos t}\right)\,dt$$ $$\Rightarrow I=\frac{\pi}{2}-\frac{1}{2}\int_0^\frac{\pi}{2}\frac{1}{\cos^2\frac{t}{2}}\,dt=\frac{\pi}{2}-1$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.