Is Trig Substitution Necessary for Solving This Integral?

[PCSL]

first off, i can solve this problem easily with u-sub but the question asks to use trig sub.

[itex]\int[/itex]4x/(x²+1)

then i have x=tan(theta)
dx=sec²(theta)

=[itex]\int[/itex]4tan(theta)sec²(theta)/(tan²(theta)+1)

=[itex]\int[/itex]4tan(theta)d(theta)

=-4ln(cos(theta))=-4ln(cos(arctan(x)))

thanks for the help, this problem is driving me crazy since it appears so simple.

edit: The answer is 2ln(x²+1)+C

 

[Char. Limit]

What you really need to do here is draw a triangle. Say that you draw a triangle with the 90-degree corner on the bottom right. Now we know that tan(θ)=y/x, where y is the vertical leg of the triangle and x is the horizontal length. So if we let the vertical length of the triangle be x, and the horizontal length 1, what does this make the hypotenuse? And then, if you know the adjacent side and the hypotenuse, you can figure out cos(θ), and then just remember that θ=arctan(x).

 

[PCSL]

What you really need to do here is draw a triangle. Say that you draw a triangle with the 90-degree corner on the bottom right. Now we know that tan(θ)=y/x, where y is the vertical leg of the triangle and x is the horizontal length. So if we let the vertical length of the triangle be x, and the horizontal length 1, what does this make the hypotenuse? And then, if you know the adjacent side and the hypotenuse, you can figure out cos(θ), and then just remember that θ=arctan(x).

The hypotenuse would be rad(x²+1). I am pretty sure, however, that I messed up my integration. Could you check that since I come up with -4ln(1/rad(x²+1)) instead of 2ln(x²+1).

 

[Char. Limit]

Assuming that rad(z) is equal to the square root of z, the two answers are actually equivalent. You just need to remember a few properties of logarithms:

[tex]-4 ln\left(\frac{1}{\sqrt{x^2+1}}\right) = 4 ln\left(\sqrt{x^2+1}\right) = 2 ln\left(x^2+1\right)[/tex]

I move from the first expression to the second (and from the second to the third) by using the logarithmic law a*ln(x)=ln(x^a). In the first case, a=-1, and in the second case a=2.

 

[PCSL]

Assuming that rad(z) is equal to the square root of z, the two answers are actually equivalent. You just need to remember a few properties of logarithms:

[tex]-4 ln\left(\frac{1}{\sqrt{x^2+1}}\right) = 4 ln\left(\sqrt{x^2+1}\right) = 2 ln\left(x^2+1\right)[/tex]

I move from the first expression to the second (and from the second to the third) by using the logarithmic law a*ln(x)=ln(x^a). In the first case, a=-1, and in the second case a=2.

I could have looked at that the rest of my life and never realized they were equal. Thank you.

 

[HallsofIvy]

However, unless you are specifically required to, you should NOT use a trig substitution for this problem. The substitution [itex]u= x^2+ 1[/itex] is much simpler.

And, by the way, never write an integral without the correct differential- your problem is NOT "[itex]\int 4x/(x^2+ 1)[/itex]", it is [itex]\int [4x/(x^2+ 1)] dx[/itex].

 

[PCSL]

However, unless you are specifically required to, you should NOT use a trig substitution for this problem. The substitution [itex]u= x^2+ 1[/itex] is much simpler.

And, by the way, never write an integral without the correct differential- your problem is NOT "[itex]\int 4x/(x^2+ 1)[/itex]", it is [itex]\int [4x/(x^2+ 1)] dx[/itex].

We were required to do trig sub for this problem. And thanks, I often forget that.