1
$\begingroup$

I have this definite integral:

$$ \int_0^\Pi \cos{x} \sqrt{\cos{x}+1} \, dx $$

For finding the indefinite integral, I have tried substitution, integration by parts, but I'm having trouble solving it.

By parts

$$ \int \cos{x} \sqrt{\cos{x}+1} \, dx\ = \sqrt{\cos{x}+1}\sin{x} + \frac{1}{2} \int \frac{\sin^{2}{x}}{\sqrt{\cos{x}+1}} \, dx $$

$ f(x) = \sqrt{\cos{x}+1} \\ f'(x) = \frac{1}{2} \frac{-\sin{x}}{\sqrt{\cos{x}+1}} \\ g(x) = \sin{x} \\ g'(x) = \cos{x} $

I don't know how to approach this further because of the $\sin^{2}{x}$.

Substitution

$ \cos{x} + 1 = u \\ -\sin{x} \, dx = du $

But I have no use for $sin\,x$.

I believe it has something to do with trig manipulations.

WolframAlpha tells me to substitute, but I don't understand how to get the first u-substituted integral like shown:

enter image description here

I would really appreciate any help on this. Thank you.

$\endgroup$
  • $\begingroup$ Thank you @heropup - very well explained solution. $\endgroup$ – Kristaps Folkmanis Jan 5 '14 at 12:07
  • $\begingroup$ BTW I would say that $\pi$ $\pi$ is more common notation than $\Pi$ $\Pi$. $\endgroup$ – Martin Sleziak Jan 5 '14 at 12:52
1
$\begingroup$

Here is how I would do it: first, let's recall the cosine double-angle identity. $$\cos 2x = \cos^2 x - \sin^2 x = \cos^2 x - (1 - \cos^2 x) = 2\cos^2 x - 1.$$ Thus the corresponding half-angle identity can be written $$\cos x = \sqrt{\frac{1 + \cos 2x}{2}}$$ or equivalently, $$\sqrt{1 + \cos x} = \sqrt{2} \cos \frac{x}{2}, \quad 0 \le x \le \pi.$$ So the integral becomes $$I = \int_{x=0}^\pi \sqrt{2} \cos x \cos \frac{x}{2} \, dx.$$ Now recall the angle addition identity $$\cos(a \pm b) = \cos a \cos b \mp \sin a \sin b,$$ from which we obtain $$\cos (a+b) + \cos (a-b) = 2 \cos a \cos b.$$ Then with $a = x$, $b = x/2$, we easily see the integral is now $$I = \frac{1}{\sqrt{2}} \int_{x=0}^\pi \cos \frac{3x}{2} + \cos \frac{x}{2} \, dx.$$ Now it is a simple matter to integrate each term: $$\begin{align*} I &= \frac{1}{\sqrt{2}} \left[ \frac{2}{3} \sin \frac{3x}{2} + 2 \sin \frac{x}{2} \right]_{x=0}^\pi \\ &= \frac{1}{\sqrt{2}} \left( -\frac{2}{3} + 2 \right) = \frac{2\sqrt{2}}{3}. \end{align*} $$

$\endgroup$
1
$\begingroup$

HINT: Note that the sine function is nonnegative on the interval of integration $[0,\pi]$; that is, for all $0\leq x \leq \pi$, $\sin{x}=|\sin{x}|$. If you are substituting $u=\cos{x}+1$,

$$u=\cos{x}+1\\ \Leftrightarrow u-1=\cos{x}\\ \Leftrightarrow \left(u-1\right)^2=\cos^2{x}\\ \Leftrightarrow -\cos^2{x}=-\left(u-1\right)^2\\ \Leftrightarrow 1-\cos^2{x}=1-\left(u-1\right)^2\\ \Leftrightarrow \sin^2{x}=u\left(2-u\right)\\ \Leftrightarrow |\sin{x}|=\sqrt{u}\sqrt{2-u}.$$

$\endgroup$
  • $\begingroup$ I like your approach and I can see how this would be useful. Though, further we have to take derivatives from both sides, right? $\endgroup$ – Kristaps Folkmanis Jan 5 '14 at 12:10
  • $\begingroup$ @KristapsFolkmanis That depends on what you're trying to accomplish. In your question, you said you tried the substitution $u=\cos{x}+1$, and you correctly took the derivative of both sides to get $du$ written as a function purely of $x$ times $dx$: $du=-\sin{x}dx$. But then you said you didn't know what to do with the $\sin{x}$. My hint was supposed to help you solve for $dx$ written as a function purely of $u$ times $du$: $dx=-\frac{du}{\sin{x}}=-\frac{du}{\sqrt{u}\sqrt{2-u}}$. $\endgroup$ – David H Jan 5 '14 at 12:51
1
$\begingroup$

Note $\displaystyle \sqrt{\cos x+1}=\sqrt{2\cos^2\frac x2-1+1}=\sqrt2\left|\cos \frac x2\right|$

We have

$$\displaystyle \int_0^\pi \cos x \sqrt{\cos x+1} \, dx=\sqrt2\int_0^\pi \left(1-2\sin^2{\frac x2}\right) \cos\frac x2 \, dx$$ $$=2 \sqrt2\int_0^\pi \left(1-2\sin^2{\frac x2}\right) \, d\left(\sin\frac x2\right)$$

$\endgroup$
  • $\begingroup$ @nidia-liza, this is nice. I'd like to format if allowed $\endgroup$ – lab bhattacharjee Jan 5 '14 at 16:01
  • $\begingroup$ @lab bhattacharjee, Okey.But I do not know how to allow formatting. I almost do not speak English $\endgroup$ – nadia-liza Jan 5 '14 at 19:43
  • $\begingroup$ I've formatted the answer $\endgroup$ – lab bhattacharjee Jan 7 '14 at 15:52
0
$\begingroup$

You can also use Weierstrass substitution (tangent half-angle substitution).

$\endgroup$
-1
$\begingroup$

HINT: Try substitution $\cos x= t$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.