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How to solve $$\int \cos x \sqrt{1 - 2 \sin x} dx$$ using $u=\cos x$ as substitution?

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3 Answers 3

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Haha, funny question!

Well, the easiest way is just apply $ y = 1 - 2 \sin x $, but if you insists

Let $ u = \cos x \Rightarrow \frac {\mathrm{d}u}{\mathrm{d}x} = - \sin x $, it becomes

$ \begin{eqnarray} \displaystyle \int u \sqrt{ 1 - 2\sqrt{1-u^2}} \cdot \frac { -\mathrm{d}u}{ \sqrt{1-u^2} } \\ \end{eqnarray} $

Now let $ y = \sqrt{1-u^2} $

$ \begin{eqnarray} \displaystyle \int u \sqrt{1-2y} \cdot \frac {-1}{y} \cdot - \frac {y}{u} \mathrm{d}y \\ \end{eqnarray} $

Which equals to

$ \begin{eqnarray} \int \sqrt{1-2y} \space \mathrm{d}y & = & -\frac {1}{2} \cdot \frac {2}{3} \left (1 - 2y \right )^{3/2} + C \\ & = & { \color{red} {-\frac {1}{3} \left (1 - 2\sin x \right )^{3/2} + C} }\\ \end{eqnarray} $

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  • $\begingroup$ you actually took $y=\sqrt{1-u^2}=\sqrt{1-\cos^2x}=\sin x$ $\endgroup$
    – RE60K
    Mar 8, 2015 at 7:41
  • $\begingroup$ Yeah, but OP insists of starting with $ \cos x $! $\endgroup$
    – GohP.iHan
    Mar 8, 2015 at 7:42
  • $\begingroup$ @GohP.iHan alright, thank you! but where does the -(y/u)dy come from? $\endgroup$
    – user131398
    Mar 8, 2015 at 8:08
  • $\begingroup$ differentitate $ y = \sqrt{1-u^2} $ w.r.t. $u$ $\endgroup$
    – GohP.iHan
    Mar 8, 2015 at 8:09
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Without any substitution:

$$\int\cos\sqrt{1-2\sin x}dx=-\frac12\int(1-2\sin x)'\sqrt{1-2\sin x}dx=$$

$$=-\frac12\frac23(1-2\sin x)^{3/2}+C=-\frac13(1-2\sin x)^{3/2}+C$$

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  • $\begingroup$ Apparently someone didn't like my answer, no matter whether it is correct or not or whether it actually helps the OP or nopt. Oh, well. $\endgroup$
    – Timbuc
    Mar 8, 2015 at 7:29
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Substitute $u=1-2\sin x$ to get$$\int \cos x\sqrt{1-2\sin x}\, dx=-\frac 12\int \sqrt{u}\,du$$ Can you take it from there?

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  • $\begingroup$ same as @timbuc. $\endgroup$
    – RE60K
    Mar 8, 2015 at 7:40

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