Alas! It's an Integral

Calculus Level 2

$\int_0^{2\pi}\dfrac{\cos x}{\sqrt{4+3\sin x}}\text dx \ = \ ?$

The answer is 0.

1 solution

A Former Brilliant Member
Apr 7, 2016

Using,
$\begin{aligned} \displaystyle \int_{0}^{2a} f(x)dx &= \int_{0}^{a} f(x) + f(2a-x)dx \\ \therefore I &= \displaystyle \int_{0}^{\pi} \dfrac{\cos(x)}{\sqrt{4+3\sin(x)}}dx + \int_{0}^{\pi} \dfrac{\cos(x)}{\sqrt{4-3\sin(x)}}dx \end{aligned}$

Using that property again on the two integrals,
$\begin{aligned} I &= \displaystyle \int_{0}^{\frac{\pi}{2}} \dfrac{\cos(x)}{\sqrt{4+3\sin(x)}}dx + \int_{0}^{\frac{\pi}{2}}\dfrac{-\cos(x)}{\sqrt{4+3\sin(x)}}dx \\ &+ \int_{0}^{\frac{\pi}{2}} \dfrac{\cos(x)}{\sqrt{4-3\sin(x)}}dx + \int_{0}^{\frac{\pi}{2}}\dfrac{-\cos(x)}{\sqrt{4-3\sin(x)}}dx = 0 \end{aligned}$

Nice solution. We could also use the substitution method with $u = 4 + 3\sin(x) \to du = 3\cos(x) dx$ .

Brian Charlesworth - 5 years, 2 months ago

Exactly just a substitution $\sin x=t$ does the job and the integral transforms to:

$\displaystyle \int_{\color{#D61F06}{0}}^{\color{#D61F06}{0}}\dfrac{\mathrm{d}t}{\sqrt{4+3t}}$ which is obviously zero!!

Rishabh Jain - 5 years, 2 months ago

Same way..(+1)!

Harsh Khatri - 5 years, 2 months ago

@Harsh Khatri – :-)...... Yet again.... We often do a lot of problem 'the same way'....

Rishabh Jain - 5 years, 2 months ago

@Rishabh Jain – Exactly!

BTW nice :-P

Harsh Khatri - 5 years, 2 months ago

Hmmmm your reasoning is not correct. You would have gotten $\displaystyle \int_0^\pi \tan x \, dx= 0$ by using the substitution $y = \sin x$ .

Pi Han Goh - 5 years, 2 months ago

@Pi Han Goh – No,he is correct.The integral which you are talking about that is tan(x) has a point of discontinuity at x=pi/2.So you must split the limits from 0 to pi/2 then from pi/2 to pi. But the integral which Rishabh has solved is continuous in (0,pi).So,the process in which he has solved the integral is valid.

Indraneel Mukhopadhyaya - 5 years, 2 months ago

@Indraneel Mukhopadhyaya – Yep... .........

Rishabh Jain - 5 years, 2 months ago

@Indraneel Mukhopadhyaya – Wrong. By your logic, $\int_0^\pi \sin x \, dx = 0$ using the substitution $y= \sin x$ .

Pi Han Goh - 5 years, 1 month ago

@Pi Han Goh – How?The limits after substitution become 1 to -1.

Indraneel Mukhopadhyaya - 5 years, 1 month ago

@Indraneel Mukhopadhyaya – If you were to use the substitution $u = \sin(x)$ the limits after substitution would be $\sin(0) = 0$ and $\sin(\pi) = 0$ . then $\int_{0}^{\pi} \sin(x) dx$ would transform to

$\displaystyle\int_{0}^{0} \dfrac{1}{\sqrt{1 - u^{2}}} du = 0$ , instead of the correct value (for the original integral) of $2$ .

The problem with substitution here is that, while the limits of the transformed integral may go from $0$ to $0$ , $u$ in facttakes on the value $1$ as you go from limit to limit, thus making the integrand $\to \infty$ along the way. With the $\tan(x)$ integral Pi Han Goh mentioned before the discontinuity is immediately apparent, while here the "discontinuity" comes after the substitution.

Brian Charlesworth - 5 years, 1 month ago

I didn't want to do the normal substitution. :P

A Former Brilliant Member - 5 years, 2 months ago

Alas! It's an Integral

The answer is 0.

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