9
$\begingroup$

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

$\endgroup$
  • 3
    $\begingroup$ 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". $\endgroup$ – nbubis May 4 '14 at 4:38
  • $\begingroup$ Try to use the change of variables $x=\arctan(t/2)$. $\endgroup$ – Mhenni Benghorbal May 4 '14 at 4:40
  • 2
    $\begingroup$ @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. $\endgroup$ – Michael Hardy May 4 '14 at 5:00
7
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

HINT:

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

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

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

Use Weierstrass substitution

Can you take it home form here?

$\endgroup$

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.