What is #f(x) = int e^xcosx-secxtan^2x+sinx dx# if #f(pi/6) = 1 #?

Question

What is #f(x) = int e^xcosx-secxtan^2x+sinx    dx# if #f(pi/6) = 1   #?

Emily Ammerman · Accepted Answer

#(e^xcos(x)+e^xsin(x))/2 - (sec(x)tan(x)-ln(|tan(x)+sec(x)|))/2+cos(x)-0.96033328054#

So the first thing I'm going to do is split up the integral into three separate integrals so we can manage them a bit easier.

Solving the First:

The first one i'm looking at is

#inte^xcos(x)dx#

The only way to really do this one is by parts, which states:

#intf(x)g'(x)dx=f(x)g(x)-intf'(x)g(x)dx#

For simplicity on my part, I'm going to keep taking the derivative of cos(x) and integrating #e^x#.

So we can say

#inte^xcos(x)dx=e^xcos(x)+inte^xsin(x)dx#

I simply took out the negative from the last integral and brought it back out. We can go one step further and say:

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

And we can apply simple algebra to get the integral on the same side.

#2inte^xcos(x)dx=e^xcos(x)+e^xsin(x)#

And finally simplifying it to:

#inte^xcos(x)dx=(e^xcos(x)+e^xsin(x))/2 + C#

Solving the Second

#-intsecxtan^2xdx#

This one is a bit more complex, since it uses something called the "Reduction Formula."

The first step is going to be making everything in the same trigonometric function, which can be done by applying the following:

#tan^2(x)=sec^2(x)-1#

This simplifies our original function to:

#-intsec^3(x)-sec(x)dx#

Here's where the reduction formula is used. It's like magic: It just works:

#intsec^n(x)dx=(n-2)/(n-1)intsec^(n-2)(x)dx+(sec^(n-2)(x)tan(x))/(n-1)#

And since n is 3 in this case, it can be simplified to

#intsec^3(x)dx=(1)/(2)intsec(x)dx+(sec(x)tan(x))/2#

And we can't forget about that other segment of

#intsec(x)dx#

Luckily this is a simple integration which should probably be memorized:

#intsec(x)dx=ln(tan(x)+sec(x))+C#

By using everything we found so far, we can say the second section is equal to

#(sec(x)tan(x)-ln(|tan(x)+sec(x)|))/2+C#

Solving the Third

#intsin(x)dx=cos(x)+C#

The last one is indeed that simple. Nice way to end it off. Now, though, we can bring it all together!

#inte^xcos(x)-secxtan^2x+sinxdx# #=(e^xcos(x)+e^xsin(x))/2 - (sec(x)tan(x)-ln(|tan(x)+sec(x)|))/2+cos(x)+C#

But we're not done QUITE yet. You want an exact answer, without the +C at the end. And we know this because you're given a point on the graph. So let's just solve for C.

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

All that math adds up to 1.96033328054. So subtract that from both sides, and you have your answer of C = -0.96033328054.

Finally, you have your function:

#(e^xcos(x)+e^xsin(x))/2 - (sec(x)tan(x)-ln(|tan(x)+sec(x)|))/2+cos(x)-0.96033328054#

Ezekiel Alexander · Answer

undefined The integral of $e^x \cos x - \sec x \tan^2 x + \sin x $ with respect to $x$ is $e^x \sin x + C$, where $C$ is the constant of integration. Given that $f(\frac{\pi}{6}) = 1$, we can find the value of $C$ by substituting $x = \frac{\pi}{6}$ into the integral expression and setting it equal to 1:

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

$$ \frac{\sqrt{3}}{2} e^{\frac{\pi}{6}} + C = 1 $$

$$ \frac{\sqrt{3}}{2} \cdot \frac{e^{\pi/6}}{2} + C = 1 $$

$$ \frac{\sqrt{3}}{4} e^{\frac{\pi}{6}} + C = 1 $$

$$ C = 1 - \frac{\sqrt{3}}{4} e^{\frac{\pi}{6}} $$

Therefore, the function $f(x) = e^x \sin x - \frac{\sqrt{3}}{4} e^{\frac{\pi}{6}} + C $.