1. Trig Substitution Problems

I have 3 problems which I can't really wrap my head around. I guess the ones I've been doing so far were much more literal. I was wondering if someone could help me out

1.

2.

3. A charged rod of length L produces an electric field at point P(a,b) given by the integral below, where λ is the charge density per unit length on the rod and ε0 is the free space permittivity (see the figure) and b ≠ 0. Evaluate the integral to determine an expression for the electric field E(P). (Use lambda for λ and Emf_0 for ε0.)

2. Originally Posted by elven06
I have 3 problems which I can't really wrap my head around. I guess the ones I've been doing so far were much more literal. I was wondering if someone could help me out

1.

2.

3. A charged rod of length L produces an electric field at point P(a,b) given by the integral below, where λ is the charge density per unit length on the rod and ε0 is the free space permittivity (see the figure) and b ≠ 0. Evaluate the integral to determine an expression for the electric field E(P). (Use lambda for λ and Emf_0 for ε0.)

id help out but after trying the first one and seeing how many steps it'd take, and noticing that i still didnt match wolfram...very close to it though (so many steps!), makes me have to pass. i got quite far using integration by parts so that if $\displaystyle \int f(x)g(x)dx$ is equal to your first intergral, set f(x)=1, and take its antiderivative in the I.B.P framework. very ugly and pointless integrals.

3. Originally Posted by yoman360
The first one is correct, the second isn't. Anyone else have any input they could provide?

Thanks

4. Originally Posted by elven06
I have 3 problems which I can't really wrap my head around. I guess the ones I've been doing so far were much more literal. I was wondering if someone could help me out

1.

2.

3. A charged rod of length L produces an electric field at point P(a,b) given by the integral below, where λ is the charge density per unit length on the rod and ε0 is the free space permittivity (see the figure) and b ≠ 0. Evaluate the integral to determine an expression for the electric field E(P). (Use lambda for λ and Emf_0 for ε0.)

For 1 & 2, complete the square then use a trigonometric substitution.

5. Originally Posted by elven06
2.
$\displaystyle \int{\frac{1}{\sqrt{t^2 - 8t + 41}}\,dt} = \int{\frac{1}{\sqrt{t^2 - 8t + (-4)^2 - (-4)^2 + 41}}\,dt}$

$\displaystyle = \int{\frac{1}{\sqrt{(t - 4)^2 + 25}}\,dt}$.

Now you actually use a HYPERBOLIC substitution.

Let $\displaystyle t - 4 = 5\sinh{x}$ so that $\displaystyle dt = 5\cosh{x}\,dx$.

Also note that $\displaystyle x = \sinh^{-1}{\frac{t - 4}{5}}$.

So the integral becomes

$\displaystyle \int{\frac{1}{\sqrt{(5\sinh{x})^2 + 25}}\,5\cosh{x}\,dx}$

$\displaystyle = \int{\frac{5\cosh{x}}{\sqrt{25\sinh^2{x} + 25}}\,dx}$

$\displaystyle = \int{\frac{5\cosh{x}}{\sqrt{25(\sinh^2{x} + 1)}}\,dx}$

$\displaystyle = \int{\frac{5\cosh{x}}{5\sqrt{\cosh^2{x}}}\,dx}$

$\displaystyle = \int{\frac{\cosh{x}}{\cosh{x}}\,dx}$

$\displaystyle = \int{1\,dx}$

$\displaystyle = x + C$

$\displaystyle = \sinh^{-1}{\frac{t - 4}{5}} + C$.

Now convert to its logarithmic equivalent if need be and substitute your terminals.