What is #f(x) = int e^xcosx-tan^3x+sinx dx# if #f(pi/6) = 1 #?

Question

What is #f(x) = int e^xcosx-tan^3x+sinx    dx# if #f(pi/6) = 1   #?

Christopher Allers · Accepted Answer

#e^x/2(sin(x)+cos(x))-ln|cos(x)|-1/2sec^2(x)-cos(x)+5/3+sqrt3/2-(1/4+sqrt3/4)e^(pi/6)+ln(sqrt3/2)#

We begin by splitting the integral into three: #int\ e^xcos(x)\ dx-int\ tan^3(x)\ dx+int\ sin(x)\ dx=#

#=int\ e^xcos(x)\ dx-int\ tan^3(x)\ dx-cos(x)#

I will call the left integral Integral 1 and the right one Integral 2

Integral 1 Here we need integration by parts and a little trick. The formula for integration by parts is: #int\ f(x)g'(x)\ dx=f(x)g(x)-int\ f'(x)g(x)\ dx#

In this case, I'll let #f(x)=e^x# and #g'(x)=cos(x)#. We get that #f'(x)=e^x# and #g(x)=sin(x)#.

This makes our integral: #int\ e^xcos(x)\ dx=e^xsin(x)-int\ e^xsin(x)\ dx#

Now we can apply integration by parts again, but this time with #g'(x)=sin(x)#: #int\ e^xcos(x)\ dx=e^xsin(x)-(-e^xcos(x)-(-int\ e^xcos(x)\ dx))#

#int\ e^xcos(x)\ dx=e^xsin(x)+e^xcos(x)-int\ e^xcos(x)\ dx#

Now we can add the integral to both sides, giving: #2int\ e^xcos(x)\ dx=e^xsin(x)+e^xcos(x)#

#int\ e^xcos(x)\ dx=1/2(e^xsin(x)+e^xcos(x))+C=#

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

Integral 2 We can first use the identity: #tan(theta)=sin(theta)/cos(theta)#

This gives: #int\ tan^3(x)\ dx=int\ sin^3(x)/cos^3(x)\ dx=int\ (sin(x)sin^2(x))/cos^3(x)\ dx#

Now we can use the pythagorean identity: #sin^2(theta)=1-cos^2(theta)#

#int\ (sin(x)(1-cos^2(x)))/cos^3(x)\ dx#

Now we can introduce a u-substitution with #u=cos(x)#. We then divide by the derivative, #-sin(x)# to integrate with respect to #u#: #-int\ (cancel(sin(x))(1-cos^2(x)))/(cancel(sin(x))cos^3(x))\ du=-int\ (1-u^2)/u^3\ du=int\ u^2/u^3-1/u^3\ du=#

#=int\ 1/u-1/u^3\ du=ln|u|+1/(2u^2)+C=ln|cos(x)|+1/(2cos^2(x))+C#

Completing the original integral Now that we know Integral 1 and Integral 2, we can plug them back into the original integral and simplify to get the final answer: #e^x/2(sin(x)+cos(x))-ln|cos(x)|-1/2sec^2(x)-cos(x)+C#

Now that we know the antiderivative, we can solve for the constant: #f(pi/6)=1#

#e^(pi/6)/2(sin(pi/6)+cos(pi/6))-ln|cos(pi/6)|-1/2sec^2(pi/6)-cos(pi/6)+C=1#

#-2/3-sqrt(3)/2+1/2(1/2+sqrt(3)/2)e^(pi/6)-ln(sqrt(3)/2)+C=1#

#C=1+2/3+sqrt3/2-(1/4+sqrt3/4)e^(pi/6)+ln(sqrt3/2)#

#C=5/3+sqrt3/2-(1/4+sqrt3/4)e^(pi/6)+ln(sqrt3/2)#

This gives that our function is: #e^x/2(sin(x)+cos(x))-ln|cos(x)|-1/2sec^2(x)-cos(x)+5/3+sqrt3/2-(1/4+sqrt3/4)e^(pi/6)+ln(sqrt3/2)#

Ezra Adams · Answer

undefined To find the integral of the function $ f(x) = \int e^x \cos x - \tan^3 x + \sin x \, dx $ when $ f(\frac{\pi}{6}) = 1 $, we use the given information to determine the constant of integration.

Given: $ f(\frac{\pi}{6}) = 1 $

Integrate $ f(x) $ to get $ F(x) $:

$$ F(x) = \int e^x \cos x - \tan^3 x + \sin x \, dx $$

$$ F(x) = e^x \sin x + C $$

Now, plug in $ \frac{\pi}{6} $ into $ F(x) $ and solve for $ C $:

$$ 1 = e^{\frac{\pi}{6}} \sin \frac{\pi}{6} + C $$

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

$$ 1 = \frac{\sqrt{3}}{4} + C $$

$$ C = 1 - \frac{\sqrt{3}}{4} $$

So, the integral becomes:

$$ F(x) = e^x \sin x + 1 - \frac{\sqrt{3}}{4} $$