"Problem with definite integral∫0π6cos⁡x1−2sin⁡xdx" solved and explained

Maurice Maldonado

Answered question

2022-04-28

Problem with definite integral

\int_{0}^{\frac{π}{6}} \cos x \sqrt{1 - 2 \sin x} d x

Penelope Carson

Beginner2022-04-29Added 16 answers

Consider the integral

\int_{0}^{\frac{π}{6}} \cos x \sqrt{1 - 2 \sin x} d x

with the substution

u = \cos (x)

. Making the desired substitution the integral becomes

I = \int_{\frac{\sqrt{3}}{2}}^{1} u \sqrt{\frac{1 - 2 \sqrt{1 - u^{2}}}{1 - u^{2}}} d u

Now make the substitution

t = 1 - u^{2}

to obtain the integral

I = \int_{0}^{\frac{1}{2}} \sqrt{1 - 2 t} d t

Making one last change of

y = 1 - 2 t

which leads to

I = \frac{1}{2} \int_{0}^{1} \sqrt{y} d y = \frac{1}{3}

August Moore

Beginner2022-04-30Added 17 answers

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} d x = \frac{2}{3} x^{\frac{3}{2}} + C

\Rightarrow \int f^{'} (x) \sqrt{f (x)} d x

= \frac{2}{3} {(f (x))}^{\frac{3}{2}} + C

In our case, we have

\cos x = - \frac{1}{2} {(1 - 2 \sin x)}^{'}

and we thus get directly

\int_{0}^{\frac{π}{6}} \cos x \sqrt{1 - 2 \sin x} d x

= - \frac{1}{2} \frac{2}{3} {(1 - 2 \sin x)}^{\frac{3}{2}} ∣_{0}^{\frac{π}{6}}

= - \frac{1}{3} (1 - 2 \sin \frac{π}{6} - 1) = \frac{2}{3} \frac{1}{2} = \frac{1}{3}

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