What is #f(x) = int sinx-cos^2x dx# if #f(pi/6)=2 #?

Question

What is #f(x) = int sinx-cos^2x   dx# if #f(pi/6)=2  #?

Evelyn Agard · Accepted Answer

#f(x)=-[cosx+x/2+1/4*sin2x]+(48+15sqrt(3)+2pi)/24# #f(x)=int(sinx-cos^2x)dx# #f(x)=int(sinx-1/2(1+cos2x))dx=[-cosx-1/2(x+(sin2x)/2)]+c# #f(x)=[-cosx-x/2-1/4*sin2x]+c#, #to (1)# #f(pi/6)=[-cos(pi/6)-pi/12-1/4*sin(pi/3)]+c=2=>[-sqrt(3)/2-pi/12-1/4sqrt(3)/2]+c=2=>-[sqrt(3)/2+pi/12+1/4sqrt(3)/2]+c=2##=>-(12sqrt(3)+2pi+3sqrt(3))/24+c=2=>c=2+(15sqrt(3)+2pi)/24# #.c=(48+15sqrt(3)+2pi)/24#, #to (2)# From (1) and (2) #f(x)=-[cosx+x/2+1/4*sin2x]+(48+15sqrt(3)+2pi)/24#

Aurora Alvarado · Answer

undefined To find $ f(x) = \int (\sin x - \cos^2 x) \, dx $ given that $ f(\frac{\pi}{6}) = 2 $, integrate the expression $ \sin x - \cos^2 x $ with respect to $ x $ and then use the given value to determine the constant of integration.

First, integrate $ \sin x - \cos^2 x $ with respect to $ x $:

$$ \int (\sin x - \cos^2 x) \, dx = -\cos x - \frac{1}{3}\cos^3 x + C $$

Where $ C $ is the constant of integration.

Given that $ f(\frac{\pi}{6}) = 2 $, substitute $ x = \frac{\pi}{6} $ and solve for $ C $:

$$ 2 = -\cos\frac{\pi}{6} - \frac{1}{3}\cos^3\frac{\pi}{6} + C $$

$$ 2 = -\frac{\sqrt{3}}{2} - \frac{1}{3}\left(\frac{\sqrt{3}}{2}\right)^3 + C $$

$$ 2 = -\frac{\sqrt{3}}{2} - \frac{1}{3}\left(\frac{3\sqrt{3}}{8}\right) + C $$

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

$$ 2 = -\frac{4\sqrt{3} + \sqrt{3}}{8} + C $$

$$ 2 = -\frac{5\sqrt{3}}{8} + C $$

$$ C = 2 + \frac{5\sqrt{3}}{8} $$

Thus, the function $ f(x) = \int (\sin x - \cos^2 x) \, dx $ with $ f(\frac{\pi}{6}) = 2 $ is:

$$ f(x) = -\cos x - \frac{1}{3}\cos^3 x + 2 + \frac{5\sqrt{3}}{8} $$

Isaiah Allen · Answer

undefined To find the function $ f(x) = \int \sin(x) - \cos^2(x) \, dx $ when $ f(\frac{\pi}{6}) = 2 $, we need to evaluate the integral and then find the function.

Given that $ f(\frac{\pi}{6}) = 2 $, we can use this information to find the constant of integration.

First, we evaluate the integral:

$$ \int \sin(x) - \cos^2(x) \, dx $$

$$ = -\cos(x) - \frac{1}{3} \sin^3(x) + C $$

Now, we plug in $ x = \frac{\pi}{6} $ to find the constant of integration $ C $:

$$ -\cos(\frac{\pi}{6}) - \frac{1}{3} \sin^3(\frac{\pi}{6}) + C = 2 $$

$$ -\frac{\sqrt{3}}{2} - \frac{1}{3} \left(\frac{1}{2}\right)^3 + C = 2 $$

$$ -\frac{\sqrt{3}}{2} - \frac{1}{24} + C = 2 $$

$$ C = 2 + \frac{\sqrt{3}}{2} + \frac{1}{24} $$

Thus, the function $ f(x) = \int \sin(x) - \cos^2(x) \, dx $ is:

$$ f(x) = -\cos(x) - \frac{1}{3} \sin^3(x) + 2 + \frac{\sqrt{3}}{2} + \frac{1}{24} $$