1

So, for example, we have the integral: $$I = \int{\sqrt{1+\sin x}}dx$$ and using the WA method, we solve it like this: enter image description here

Can you please explain me the last step, why does the solution change due to restricted values? And how do I change it step by step?

Edit: This is the way the teacher did it:

$$I = \int{\sqrt{1+sinx}}dx$$ Since $\sin{x}=2\sin{\frac{x}{2}}\cos{\frac{x}{2}}$ $$I = \int{\sqrt{1+2\sin{\frac{x}{2}}\cos{\frac{x}{2}}}dx}=\int{|\sin{\frac{x}{2}}\cos{\frac{x}{2}}|}dx=\sqrt{2}\int{|\sin(x+\frac{\pi}{4})|dx}=$$ $$=C + \sqrt{2}\int_{-\frac{\pi}{4}}^x{|sin(t+\frac{\pi}{4})|dt=C+\sqrt{2}\int_0^{x+\frac{pi}{4}}{|\sin{z}|dz}}$$ If we say $k=[\frac{x+\frac{\pi}{4}}{\pi}]$ then $\frac{x+\frac{\pi}{4}}{\pi}=k+l$ where $0<=l<1$. That yields $x+\frac{\pi}{4}=k\pi+s$ where $0 <=s <\pi$ Thus, $$I=C+\sqrt{2}\int_0^{k\pi+s}{|\sin{z}|dz}=$$ $$=C+\sqrt{2}\int_0^{k\pi}{|\sin{z}|dz}+\int_{k\pi}^{k\pi+s}{|\sin{z}|dz}=$$ $$=C+2\sqrt{2}+\int_{k\pi}^{k\pi+s}{|\sin{z}|dz}=$$ $$=C+2\sqrt{2}+\int_0^s{\sin{z}dz}=$$ $$=C+2\sqrt{2}+1-\cos{s}=C_1+2\sqrt{2}k-\cos(x+\frac{\pi}{4}+k\pi)=$$ $$=C_1+2\sqrt{2}k-(-1)^k\cos(x+\frac{\pi}{4})$$

I would really appreciate the explanation of what is done here, including the restrictions.

A6SE
  • 1,645
  • 1
    Some online integrators like to use trig identities, particularly half angle formulas but I do not know why (maybe somebody else knows). But my question would be: Do you know how to do this integral without W.A? Because that is important. Hint: Multiply top and bottom by $\sqrt{1-sinx}$ – imranfat May 01 '15 at 15:32

1 Answers1

1

here is another way to do this. we will use the identity $$1 + \sin t = 1 + \cos(t - \pi/2) = 2\sin^2(t/2-\pi/4).$$ therefore $$\int \sqrt{1 + \sin t }\, dt = \sqrt 2\int\sin(t/2-\pi/4)\, dt=-2\sqrt 2\cos(t/2-\pi/4)+c $$

abel
  • 29,170
  • No restrictions there? – A6SE May 01 '15 at 16:05
  • @A6Tech, i don't see any. – abel May 01 '15 at 16:06
  • Can you help me analyze this integral the way it's done in my workbook? I'll post it in the edit section. – A6SE May 01 '15 at 16:14
  • I think there are some restrictions, you did it similarly like my teacher and I would appreciate really much if you saw the restrictions he pointed out. – A6SE May 01 '15 at 16:35
  • @A6Tech, the argument $1+\sin t$ in the square root is always nonnegative, so the square root makes sense for ant $t.$ at one place you have indefinite integral, and then you change it to definite integral. i don't understand what is going on there. – abel May 01 '15 at 17:15
  • Me neither, your solution makes sense, but I really can't figure out what the teacher did... – A6SE May 01 '15 at 17:23