What is #f(x) = int x-sin2x+cosx dx# if #f(pi/2)=3 #?

Question

What is #f(x) = int x-sin2x+cosx   dx# if #f(pi/2)=3 #?

Scarlett Allison · Accepted Answer

#f(x)=x^2/2+1/2cos(2x)+sin(x)+5/2-pi^2/8#

We will use the following rules:

So:

#f(x)=intx^1dx-intsin(2x)dx+intcos(x)dx#

The first and third can be found directly:

#f(x)=x^2/2-intsin(2x)dx+sin(x)#

For the remaining integral, let #u=2x# so #du=(2)dx#. Then:

#f(x)=x^2/2-1/2intsin(2x)(2)dx+sin(x)#

#f(x)=x^2/2-1/2intsin(u)du+sin(x)#

#f(x)=x^2/2-1/2(-cos(u))+sin(x)+C#

#f(x)=x^2/2+1/2cos(2x)+sin(x)+C#

We can now solve for #C# using #f(pi/2)=3#:

#3=(pi/2)^2/2+1/2cos(2*pi/2)+sin(pi/2)+C#

#3=pi^2/8+1/2(-1)+1+C#

#C=5/2-pi^2/8#

So:

#f(x)=x^2/2+1/2cos(2x)+sin(x)+5/2-pi^2/8#

William Abdalla · Answer

#f(x)=x^2/2+1/2cos2x+sinx+(20-pi^2)/8# #f(x)=intx-sin2x+cosxdx# #" "# #f(x)=intxdx-intsin2xdx+intcosxdx+C# #" "C#is a constant #" "# #f(x)=x^2/2-1/2intd(-cos2x)+intd(sinx)+C# #" "# #f(x)=x^2/2+1/2cos2x+sinx+C# #" "# #" "# Le us find #C# #" "# Given #f(pi/2)=3# #" "# #f(pi/2)=(pi/2)^2/2+1/2cos(2*pi/2)+sin(pi/2)+C# #" "# Substituting #f(pi/2)=3# #" "# #rArr3=(pi)^2/8+1/2cospi+1+C# #" "# #rArr3=(pi)^2/8-1/2+1+C# #" "# #rArr3=pi^2/8+1/2+C# #" "# #rArrC=3-1/2-pi^2/8# #" "# #rArrC=24/8-4/8-pi^2/8# #" "# #rArrC=(20-pi^2)/8# #" "# #" "# Therefore,#f(x)=x^2/2+1/2cos2x+sinx+(20-pi^2)/8#

Axel Alvarez · Answer

undefined To find $ f(x) = \int x - \sin^2(x) + \cos(x) \, dx $ given that $ f(\frac{\pi}{2}) = 3 $, we can differentiate $ f(x) $ with respect to $ x $ and solve for the constant of integration using the given initial condition.

Differentiating $ f(x) $:
$$ f'(x) = \frac{d}{dx} \left( \int x - \sin^2(x) + \cos(x) \, dx \right) $$
$$ = x - \sin^2(x) + \cos(x) $$

Now, integrate $ f'(x) $ with respect to $ x $:
$$ f(x) = \int (x - \sin^2(x) + \cos(x)) \, dx $$
$$ = \frac{x^2}{2} + x\sin(x) - \frac{x}{2} - \cos(x) + C $$

Given $ f\left(\frac{\pi}{2}\right) = 3 $:
$$ 3 = \frac{\left(\frac{\pi}{2}\right)^2}{2} + \frac{\pi}{2}\sin\left(\frac{\pi}{2}\right) - \frac{\pi}{4} - \cos\left(\frac{\pi}{2}\right) + C $$
$$ 3 = \frac{\pi^2}{8} + \frac{\pi}{2} - \frac{\pi}{4} + 1 + C $$
$$ 3 = \frac{\pi^2}{8} + \frac{3\pi}{4} + 1 + C $$
$$ C = 3 - \frac{\pi^2}{8} - \frac{3\pi}{4} - 1 $$
$$ C = 3 - \frac{\pi^2}{8} - \frac{3\pi}{4} - \frac{8}{8} $$
$$ C = 2 - \frac{\pi^2}{8} - \frac{3\pi}{4} $$

Therefore, the function $ f(x) = \int x - \sin^2(x) + \cos(x) \, dx $ with $ f\left(\frac{\pi}{2}\right) = 3 $ is:
$$ f(x) = \frac{x^2}{2} + x\sin(x) - \frac{x}{2} - \cos(x) + 2 - \frac{\pi^2}{8} - \frac{3\pi}{4} $$

Samantha Adkison · Answer

undefined To find $ f(x) = \int x - \sin^2(x) + \cos(x) \, dx $ when $ f(\frac{\pi}{2}) = 3 $, we'll first integrate the given function to obtain an expression for $ f(x) $. Then, we'll use the given value of $ f(\frac{\pi}{2}) = 3 $ to solve for the constant of integration.

Given function:
$ f(x) = \int x - \sin^2(x) + \cos(x) \, dx $

To integrate, we'll find the antiderivative of each term:
$ \int x \, dx = \frac{x^2}{2} + C_1 $
$ \int \sin^2(x) \, dx = \frac{x}{2} - \frac{\sin(2x)}{4} + C_2 $
$ \int \cos(x) \, dx = \sin(x) + C_3 $

Now, we'll combine the antiderivatives and the constant of integration:
$ f(x) = \frac{x^2}{2} + \frac{x}{2} - \frac{\sin(2x)}{4} + \sin(x) + C $

Given that $ f(\frac{\pi}{2}) = 3 $:
$ 3 = \frac{\left(\frac{\pi}{2}\right)^2}{2} + \frac{\frac{\pi}{2}}{2} - \frac{\sin(2 \cdot \frac{\pi}{2})}{4} + \sin(\frac{\pi}{2}) + C $

Solving for $ C $:
$ 3 = \frac{\pi^2}{8} + \frac{\pi}{4} - \frac{\sin(\pi)}{4} + 1 + C $

$ 3 = \frac{\pi^2}{8} + \frac{\pi}{4} - \frac{0}{4} + 1 + C $

$ 3 = \frac{\pi^2}{8} + \frac{\pi}{4} + 1 + C $

$ C = 3 - \frac{\pi^2}{8} - \frac{\pi}{4} - 1 $

$ C = 3 - \frac{\pi^2}{8} - \frac{2\pi}{8} - \frac{8}{8} $

$ C = 3 - \frac{\pi^2 + 2\pi + 8}{8} $

$ C = \frac{24 - \pi^2 - 2\pi - 8}{8} $

$ C = \frac{16 - \pi^2 - 2\pi}{8} $

$ C = \frac{16}{8} - \frac{\pi^2}{8} - \frac{2\pi}{8} $

$ C = 2 - \frac{\pi^2}{8} - \frac{\pi}{4} $

Therefore, the expression for $ f(x) $ is:
$ f(x) = \frac{x^2}{2} + \frac{x}{2} - \frac{\sin(2x)}{4} + \sin(x) + 2 - \frac{\pi^2}{8} - \frac{\pi}{4} $