# How find this integral $\int\frac{\sin{x}}{\sqrt{2}+\sin{x}+\cos{x}}dx$

Find the integral $$\int\dfrac{\sin x}{\sqrt{2}+\sin x+\cos x} \, dx$$

My idea: since $$\sin x+\cos x=\sqrt{2}\sin(x+\dfrac{\pi}{4})$$ so

$$\int\dfrac{\sin x}{\sqrt{2}+\sin x +\cos x} \, dx=\int\dfrac{\sin x}{\sqrt{2}(1+\sin (x+\dfrac{\pi}{4})} \, dx$$ But then I don't know how to continue. Thank you

• Since you seem to be using the term a lot, please be advised that "Then I can't", is not really English. You may want to replace that with something akin to "But then I don't know how to continue". – nbubis May 4 '14 at 4:38
• Try to use the change of variables $x=\arctan(t/2)$. – Mhenni Benghorbal May 4 '14 at 4:40
• @nbubis : In some contexts, "Then I can't" is perfect English. When I understand a question, then I can answer it. But jsut suppose in some particular case I don't. Then I can't. – Michael Hardy May 4 '14 at 5:00

Hint: lab bhattacharjee's Weierstrass substitution hint is very appropriate of course, but it would be nice if more of the algebra tedium could be obviated. Try,

$$\int\dfrac{\sin{x}}{\sqrt{2}+\sin{x}+\cos{x}}dx =\frac{1}{\sqrt{2}}\int\dfrac{\sin{x}}{1+\sin{(x+\dfrac{\pi}{4})}}dx\\ =\frac{1}{\sqrt{2}}\int\dfrac{\sin{(\phi-\frac{\pi}{4})}}{1+\sin{\phi}}d\phi\\ =\frac{1}{2}\int\dfrac{\sin{\phi}-\cos{\phi}}{1+\sin{\phi}}d\phi$$

Now use the Weierstrass substitution.

It is a nice idea. For no essential reason, I would prefer to use $\sin x+\cos x=\sqrt{2}\cos(x-\pi/4)$. Making the substitution $t=x-\pi/4$, and using the fact that the $\sin x$ on top is equal to $\frac{\sin t+\cos t}{\sqrt{2}}$, we arrive at the integral $$\int\frac{1}{2}\frac{\sin t+\cos t}{1+\cos t}\,dt.$$

Now note that $1+\cos t=2\cos^2(t/2)$, and express the numerator as $2\sin(t/2)\cos(t/2)+2\cos^2(t/2)-1$. The rest is downhill.

In general, Weierstrass is probably a good idea for such trigonometric integrals. However, your progress left the denominator much more manageable. I would start as in David H's answer up until the step

$$\frac12\int\frac{\sin\theta-\cos\theta}{1+\sin\theta}d\theta$$

Instead of Weierstrass from here, simply multiply by $\frac{1-\sin\theta}{1-\sin\theta}$

$$\frac12\int\frac{(\sin\theta-\cos\theta)(1-\sin\theta)}{1-\sin^2\theta}d\theta=\frac12\int\frac{\sin\theta-\cos\theta-\sin^2\theta+\sin\theta\cos\theta}{\cos^2\theta}d\theta=$$ $$\frac12\int\sec\theta\tan\theta d\theta-\frac12\int\sec\theta d\theta-\frac12\int(\sec^2\theta-1)d\theta+\frac12\int\tan\theta d\theta$$

You should have no trouble with these remaining integrals.

HINT:

Let $$\sin x=A(\sqrt2+\sin x+\cos x)-B d(\sqrt2+\sin x+\cos x)+C$$

$$\implies\sin x=\sqrt2A+C+\sin x(A+B)+\cos x(A-B)$$

$$\implies A-B=0\iff A=B,A+B=1\implies A=B=\frac12,C=-\sqrt2A=\cdots$$

Can you take it home form here?