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$$\int_{0}^{\frac{\pi}{6}}\cos x\sqrt{1-2\sin x} dx$$

The question says 'evaluate the integral using the suggested substitution. It gives $u=\cos x$. But I think Let $u=1-2\sin x$ is better.

$$\int_{0}^{\frac{\pi}{6}}\cos x\sqrt{1-2\sin x} dx$$

$$u=1-2\sin x$$

$$du=-2\cos x dx$$

$$=-\frac{1}{2}\int_{1}^{0}\sqrt{u}du$$

$$=\frac{1}{2}\int_{0}^{1}\sqrt{u}du$$

$$=\left | \frac{u^{\frac{3}{2}}}{3} \right |_{0}^{1}$$

$$=\frac{1}{3}$$

My question is how to solve it by using $u=\cos x$. Can anyone show the solution for it? Thanks a lot!

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    $\begingroup$ But it's $2\sin{x}$ not $\sin{2x}$ $\endgroup$ – Arpan Mar 16 '15 at 5:39
  • $\begingroup$ @Nilan It's not sin2x but 2sinx. Have you got any idea for it? Thanks. $\endgroup$ – Mathxx Mar 16 '15 at 5:47
  • $\begingroup$ @Mathxx Let me know if I can help further here! I just want to give you the best answer I can give you. $\endgroup$ – Mark Viola Mar 16 '15 at 6:17
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Well, @Nilan has the better way to go. But, here is another "Brute Force," double substitution method that works.

Write, $\sin x = \sqrt{1-\cos^2 x}$ (the positive square root is appropriate here since $0\le x\le \frac{\pi}{6}$. Then, with $u = \cos x$, $du = -\sin x dx=-\sqrt{1-u^2}du$, and the limits of integration extend from $u=1$ to $u=\frac{\sqrt{3}}{2}$. Thus,

$$\int_0^{\frac{\pi}{6}}\cos x\sqrt{1-2\sin x}dx= \int_{\frac{\sqrt{3}}{2}}^1 \frac{u\sqrt{1-2\sqrt{1-u^2}}}{\sqrt{1-u^2}}du$$

Now, we make a second substitution. Let $y=\sqrt{1-u^2}$. Then, $dy=\frac{-2u}{\sqrt{1-u^2}}du$ and the limits of integration go from $y=\frac12$ to $y=0$. (Note: This second substitution is identical to making the original substitution $u=\sin x$). Thus,

$$\int_0^{\frac{\pi}{6}}\cos x\sqrt{1-2\sin x}dx=\int_0^{\frac12} \sqrt{1-2y}dy=\frac13$$

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  • $\begingroup$ I have the same solution, but slow typing caused me to be one minute behind $\endgroup$ – Leucippus Mar 16 '15 at 6:17
  • $\begingroup$ @Leucippus That happens to me a lot too. No worry. Glad to see someone else likes the brute force approach here. $\endgroup$ – Mark Viola Mar 16 '15 at 6:19
  • $\begingroup$ @Dr.MV. Why your lower bound is $\frac{\sqrt{3}}{2}$ instead of 1? Thanks. $\endgroup$ – Mathxx Mar 16 '15 at 6:20
  • $\begingroup$ I absorbed the minus sign in the new differential and flipped the integration limits. $\endgroup$ – Mark Viola Mar 16 '15 at 6:22
  • $\begingroup$ @Dr.MV. Yea I've understood. Wonder why I'm so stupid :( $\endgroup$ – Mathxx Mar 16 '15 at 6:29
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Consider the integral \begin{align} \int_{0}^{\frac{\pi}{6}}\cos x\sqrt{1-2\sin x} dx \end{align} with the substitution $u = \cos(x)$. Making the desired substitution the integral becomes \begin{align} I = \int_{\sqrt{3}/2}^{1} u \sqrt{ \frac{1 - 2 \sqrt{1-u^{2}}}{1 - u^{2} } } du \end{align} Now make the substitution $t = 1 - u^{2}$ to obtain the integral \begin{align} I = \int_{0}^{1/2} \sqrt{1 - 2t} \, dt. \end{align} Making one last change of $y = 1-2t$ which leads to \begin{align} I = \frac{1}{2} \int_{0}^{1} \sqrt{y} \, dy = \frac{1}{3}. \end{align}

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The substitution is fine, and if you were directed to do so then there is no option, but I think there's a simpler way to go. First, observe that

$$\int\sqrt x\;dx=\frac23 x^{3/2}+C\implies \int f'(x)\sqrt{f(x)}\;dx=\frac23(f(x))^{3/2}+C$$

In our case, we have

$$\cos x=-\frac12(1-2\sin x)'$$

and we thus get directly

$$\int_0^{\pi/6}\cos x\sqrt{1-2\sin x}\;dx=\left.-\frac12\frac23(1-2\sin x)^{3/2}\right|_0^{\pi/6}=$$

$$=-\frac13\left(1-2\sin\frac\pi6-1\right)=\frac23\frac12=\frac13$$

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