Calculus problem #2196

The thing we have to do here is to solve it for integration by parts.

The Equation for integration by parts goes like this:

$\int_0^\pi \! f(x)g'(x)\, \mathrm{d}x = f(x)g(x) - \int_0^\pi \! f'(x)g(x)\, \mathrm{d}x$

So we have to apply this rule where:

$f(x) = x$

… and …

$g(x) = (\sin x + \cos x)$

Because of this we can say that the integral is equivalent to the following function:

-It's important to remember that the pi and the 0 represent radian values in terms of the trigonometric functions.

$\int_0^\pi x(\sin x + \cos x) dx = x(\sin x - \cos x) - \int_0^\pi (\sin x + \cos x) dx$

If you use basic integration you'll find that:

$\int_0^\pi (\sin x + \cos x) dx = 2$

So:

$= x\sin x - x\cos x - 2$

There for if we take the original integral we have to find the following:

$= x\sin x - x\cos x - 2$ … as an integral from 0 to $\pi$ radians

So we have the following equation:

$= {[(\pi)\sin(\pi) - (\pi)\cos (\pi)] - [(0)\sin (0) - (0)\cos (0)]} -2$

$= -\pi\cos (\pi) - 2$

$= -\pi(-1) - 2$

$= \pi -2$

$= 3.1415926 - 2$

$= 1.1415926$

The answer thus would be : $\boxed{1.1415926}$

This is how you can work out the answer.

This integral can easily be evaluated by splitting it into two integrals and using integration by parts on the two integrals. Alternatively, one can use the identity $$\sin x+\cos x=\sqrt{2} \sin \left(x+\frac{\pi}{4}\right)$$ to necessitate the use of integration by parts once. Using this identity, we get $$\sqrt{2}\int 0^{\pi} x\left( \sin \left(x+\frac{\pi}{4}\right) \right) \, dx.$$ Using integration by parts with $$u=x \Rightarrow du=dx,$$ $$dv= \sin \left(x+\frac{\pi}{4}\right) \Rightarrow v= -\cos \left(x+\frac{\pi}{4}\right)$$ gives us $$\sqrt{2} \left( \left[ -x \cos \left(x+\frac{\pi}{4}\right) \right] 0^{\pi} + \int 0^{\pi} \cos \left(x+\frac{\pi}{4}\right) \, dx \right)$$ $$=\sqrt{2} \left( \frac{\pi}{\sqrt{2}}+\left[ \sin \left(x+\frac{\pi}{4}\right) \right] 0^{\pi} \right)$$ $$=\sqrt{2}\left(\frac{\pi}{\sqrt{2}} - \sqrt{2} \right) = \boxed{\pi-2}.$$