Cody's calculus question

We use the property

$\LARGE \int_0^{2a} f(x)\mathrm{d}x=2 \int_0^{a} f(x)\mathrm{d}x$

This is true when the function is even about $a$ i.e.

$\LARGE f(a+x)=f(a-x)$ $x\in [0,a]$

So, our integral becomes

$\LARGE 2\int_0^\frac{\pi}{2} \sqrt{1+\sin{x}}\mathrm{d}x$

We also use the property that

$\LARGE \int_a^{b} f(x)\mathrm{d}x=\int_a^{b} f(a+b-x)\mathrm{d}x$

Now we can write the integral as

$\LARGE 2\int_0^\frac{\pi}{2} \sqrt{1+\sin({\frac{\pi}{2}-x})}\mathrm{d}x$

Now, we use

$\LARGE \sin ({\frac{\pi}{2}-x})=\cos x$

And

$\LARGE 1+\cos x=2\cos^{2}\frac{x}{2}$

Now, our integral becomes

$\LARGE 2\sqrt{2}\int_0^\frac{\pi}{2} \cos(\frac{x}{2})\mathrm{d}x$

$\LARGE \Rightarrow4 \sqrt{2} \sin \frac{x}{2}\mid_0^\frac{\pi}{2}$

$\LARGE \Rightarrow 4$

But the question requires us to find $100$ times the integral.

So the answer is $\boxed{400}$

We know that if $f(x) = f(2a -x)$ ,

$\displaystyle \int_{0}^{2a} f(x) dx = 2\int_{0}^{a} f(x) dx$

Hence, $\displaystyle \int_{0}^{\pi} \sqrt{1 + \sin x} dx$

= $2 \displaystyle \int_{0}^{\frac{\pi}{2}} \sqrt{1+\sin x} dx$

Substituting $x = \sin^{-1} t,$

we get $dx = \frac{dt}{\sqrt{1-t^2}}$

Hence we get $\displaystyle \int_{0}^{1} 2\frac{\sqrt{1+t}}{\sqrt{1-t^2}} dt = 2\int_{0}^{1} \frac{dt}{\sqrt{1-t}} = 4$

Hence, multiplying by $100$ , we get $400$

(1+sinx)^1/2=(sin^2x/2+cos^x/2+2sinx/2*cosx/2)^2 =[(sinx/2+cosx/2)^2]^1/2 =sinx/2+cosx/2 then by applying integral on both sides ANS=400

Let $I=\int_0^\pi\sqrt{1+\sin x}dx$ and note that $\frac{1}{2}I=\int_0^{\pi/2}\sqrt{1+\sin x} dx=\int_{\pi/2}^\pi\sqrt{1+\sin x} dx$ by symmetry.

Then,

$\int_0^{\pi/2}\sqrt{1+\sin x} dx=\int_0^{\pi/2}\sqrt{1+\sin x}\cdot\frac{\sqrt{1-\sin x}}{\sqrt{1-\sin x}} dx=\int_0^{\pi/2} \frac{|cos x|}{\sqrt{1-\sin x}} dx.$

Since $\cos x\geq 0$ on $[0,\pi/2]$ , $|\cos x|=\cos x$ there and so the substitution $u=\sin x$ turns the above integral into $\int_0^1 (1-u)^{-1/2} du=2$ .

So we find that $I/2=2$ . This gives $I=4$ , so $100I=400$ is our answer.

Another method Substitute $\sin { \theta } =\quad \frac { 2\tan { \frac { \theta }{ 2 } } }{ 1+\tan ^{ 2 }{ \frac { \theta }{ 2 } } }$ Then \int { \sqrt { 1+\quad \frac { 2\tan { \frac { \theta }{ 2 } } }{ 1+\tan ^{ 2 }{ \frac { \theta }{ 2 } } } } } d\theta \quad =\quad \int { \frac { \sqrt { { \left( 1+\tan { \frac { \theta }{ 2 } } \right) }^{ 2 } } }{ \sec { \frac { \theta }{ 2 } } } } d\theta \\ \Longrightarrow \quad =\quad \int { \frac { 1+\tan { \frac { \theta }{ 2 } } }{ \sec { \frac { \theta }{ 2 } } } d\theta } =\quad \int { \cos { \frac { \theta }{ 2 } } +\sin { \frac { \theta }{ 2 } } } d\theta \\ \Longrightarrow \quad =\quad { 2\left[ \sin { \frac { \theta }{ 2 } } -\cos { \frac { \theta }{ 2 } } \right] }_{ 0 }^{ \Pi }\quad =\quad 4\\ Hence,\quad 100\quad \times \quad 4\quad =\quad 400

we know that sin(x)=2 sin(x/2) cos(x/2) so

1+sin(x)=1+2 sin(x/2) cos(x/2 ) = sin(x/2)^2 + cos(x/2)^2 + 2 sin(x/2) cos(x/2)

={ sin(x/2) + cos (x/2) }^2

rest we know how to proceed.......................

sqrt(1+sinx)=sqrt((sinx)^2+(cosx)^+2sinx cosx)=sin(x/2)+cos(x/2) now integrating sin(x/2)+cos(x/2) from limit 0 to pi we get 2 200*2=400 (ans)

We will begin by attempting a substitution:

$u=1+\sin{x}$

$\mathrm{d}u=\cos{x}\mathrm{d}x$

No cosine can be found in the integral so we will multiply by $\frac{\cos{x}}{\cos{x}}$ :

$100\int_0^\pi\frac{\cos{x}\sqrt{1+\sin{x}}}{\cos{x}}\,\mathrm{d}x$

Next we recall that:

$\sin^2{x}+\cos^2{x}=1$

We also note that when $0<x<\frac{\pi}{2}$ :

$cos{x}>0$

And when $\frac{\pi}{2}<x<\pi$ :

$cos{x}<0$

Thus, when $0<x<\frac{\pi}{2}$ :

$cos{x}=\sqrt{1-\sin^2{x}}$

And when $\frac{\pi}{2}<x<\pi$ :

$cos{x}=-\sqrt{1-\sin^2{x}}$

Then we split the integral accordingly and substitute:

$100\int_0^\frac{\pi}{2}\frac{\cos{x}\sqrt{1+\sin{x}}}{\cos{x}}\,\mathrm{d}x+100\int_\frac{\pi}{2}^\pi\frac{\cos{x}\sqrt{1+\sin{x}}}{\cos{x}}\,\mathrm{d}x$

$=100\int_0^\frac{\pi}{2}\frac{\cos{x}\sqrt{1+\sin{x}}}{\sqrt{1-\sin^2{x}}}\,\mathrm{d}x+100\int_\frac{\pi}{2}^\pi\frac{\cos{x}\sqrt{1+\sin{x}}}{-\sqrt{1-\sin^2{x}}}\,\mathrm{d}x$

$=100\int_1^2\frac{\sqrt{u}}{\sqrt{1-(u-1)^2}}\,\mathrm{d}u-100\int_2^1\frac{\sqrt{u}}{\sqrt{1-(u-1)^2}}\,\mathrm{d}u$

We note that by flipping the limits on the right integral, we can negate that integral and add the two integrals together:

$200\int_1^2\frac{\sqrt{u}}{\sqrt{1-(u-1)^2}}\,\mathrm{d}u$

$=200\int_1^2\frac{1}{\sqrt{2-u}}\,\mathrm{d}u$

Lastly, we can take the anti-derivative and simplify:

$200(-2\sqrt{2-u})\bigg\rvert_1^2$

$=-400(\sqrt{2-2} - \sqrt{2-1})$

$=-400(-1)$

$=\boxed{400}$