What is #f(x) = int xsin2x-6cotx dx# if #f(pi/4)=1 #?

Question

What is #f(x) = int xsin2x-6cotx   dx# if #f(pi/4)=1 #?

Charles Arocha · Accepted Answer

#f(x)=-x/2*cos2x+1/4*sin2x-6ln(sinx)+1+(pisqrt2)/16-3Ln2#

Due to #int xsin2x*dx=x*(-1/2*cos2x)-int (-1/2)*cos2x*dx#

=#-x/2*cos2x+1/2intcos2x*dx#

=#-x/2*cos2x+1/4sin2x+C#

#f(x)=int (xsin2x-6cotx)*dx=-x/2*cos2x+1/4*sin2x-6ln(sinx)+C#

After imposing #f(pi/4)=1# condition,

#-pi/8*sin(pi/4)-6Ln(sin(pi/4))+C=1#

#-(pisqrt2)/16-6Ln(sqrt2/2)+C=1#

#-(pisqrt2)/16+3Ln2+C=1#

#C=1+(pisqrt2)/16-3Ln2#

Thus,

#f(x)=-x/2*cos2x+1/4*sin2x-6ln(sinx)+1+(pisqrt2)/16-3Ln2#

Paisley Johnson · Answer

undefined To find $ f(x) = \int x\sin(2x) - 6\cot(x) \, dx $ given that $ f(\frac{\pi}{4}) = 1 $, we first need to find the antiderivative of $ x\sin(2x) - 6\cot(x) $. Then, we can use the given initial condition to solve for the constant of integration.

$$ \int x\sin(2x) - 6\cot(x) \, dx = \frac{1}{2}x(-\cos(2x)) - 6\ln|\sin(x)| + C $$

Given $ f(\frac{\pi}{4}) = 1 $, substitute $ \frac{\pi}{4} $ into the antiderivative and equate it to 1:

$$ \frac{1}{2}(\frac{\pi}{4})(-\cos(\frac{\pi}{2})) - 6\ln|\sin(\frac{\pi}{4})| + C = 1 $$

Simplify and solve for $ C $:

$$ \frac{1}{2}(\frac{\pi}{4})(0) - 6\ln|\frac{1}{\sqrt{2}}| + C = 1 $$

$$ 0 - 6\ln|\frac{1}{\sqrt{2}}| + C = 1 $$

$$ -6\ln(\sqrt{2}) + C = 1 $$

$$ C = 1 + 6\ln(\sqrt{2}) $$

So, the function $ f(x) $ is:

$$ f(x) = \frac{1}{2}x(-\cos(2x)) - 6\ln|\sin(x)| + 1 + 6\ln(\sqrt{2}) $$