# Definite integration problem (trig).

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

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

## 5 Answers

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*}

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 $u=\cos{x}+1$,

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

• I like your approach and I can see how this would be useful. Though, further we have to take derivatives from both sides, right? – Kristaps Folkmanis Jan 5 '14 at 12:10
• @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 $du$: $dx=-\frac{du}{\sin{x}}=-\frac{du}{\sqrt{u}\sqrt{2-u}}$. – David H Jan 5 '14 at 12:51

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

We have

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

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

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

HINT: Try substitution $\cos x= t$