Integration of 1/[(x^2+100)^(3/2)] - Need Help!

[StephenDoty]

The problem is $$\int$$ 1/[(x^2+100)^(3/2)] dx

u cannot equal x^2 + 100 since the derivative is 2x and there is not another x in the problem. So I am unsure of even how to start. I did it with my calculator and the answer did not seem to follow the integration by parts format. Therefore, I do not know how to start integrating this problem by hand. Any help would be greatly appreciated.

Thank you.

Stephen

 

[snipez90]

Try substituting x = 10tan(t).

 

[StephenDoty]

ok so I replace x with 10tan(t)
giving
$$\int$$ 1/[(10tan(t))^2 +100)^(3/2)] dx
thus
$$\int$$ 1/ [(100tan(t)^2 +100)^(3/2)] dx

Then say we change tan(t)^2 into sin(t)^2/cos(t)^2 making
$$\int$$ 1/ [(100(sin(t)^2/cos(t)^2) +100)^(3/2) dx
because the sin(t)^2/cos(t)^2 is to the (3/2) power I cannot use the half-angle formulas: cos(t)^2= .5(1+cos(2t)) or sin(t)^2 = .5(1-cos(2t)) to integrate.

So what do I do now?

Thank you for your help.
Stephen

 

[snipez90]

Well you have to differentiate x = 10tan(t). You've done u-substitution before right? Also note that $$100tan^2(t)+100 = 100(tan^2(t) + 1) = 100sec^2(t)$$. Anyways let me know if you have questions about the differentiation part.

 

[StephenDoty]

why would you have to derivate x=10tan(t), where dx=10sec(t)^2?
do you mean that u=10tan(x)? And if so what would the u replace?
Would you replace x with u? Making du=10sec(x)^2 dx or dx=du/10sec(t)^2?

$$\int$$ 1/ [(100tan(x)^2 +100)^(3/2)] dx

$$\int$$ 1/ [(100sec(x)^2)^(3/2)]*(10sec(x)^2) du

I am unsure as to where to put the x=10tan(t) and the derivative dx=10sec(t)^2

Thanks for the help.
Stephen

 

[Hitman2-2]

Well, you are changing the variable you are integrating with respect to, so everywhere you see x, replace it with $$10 \tan (t)$$ but don't forget about the dx. From $$x = 10 \tan (t)$$ we get $$dx = 10 \sec ^2 (t) dt$$ so replace dx with $$10 \sec ^2 (t) dt$$.

 

[StephenDoty]

so
$$\int$$ 1/[(x^2+100)^(3/2)] dx from -5 to 10
=
$$\int$$1/[((10tan(t))^2+100)^(3/2)] 10sec(t)^2 dt
=
$$\int$$1/[(100sec(t)^2)^(3/2)] * 10sec(t)^2 dt
=
$$\int$$ [10sec(t)^2]/[100sec(t)^3] dt
=
(1/10) $$\int$$ [sec(t)^2]/[sec(t)^3] dt
=
(1/10) $$\int$$ 1/ sec(t) dt
=
(1/10) $$\int$$ cos(t) dt
=
(1/10)sin(t) +c
and if we solve this from -5 to 10
wouldn't it be (1/10)sin(10) - (1/10)sin(-5)
or
would it be (1/10)sin(tan(1)^(-1)) - (1/10)sin(tan(-.5)^(-1))
since x goes from -5 to 10, using x=10tan(t), would t=tan(10/10)^(-1) and t=tan(-5/10)^(-1)?
did I do this right?
And if so why did we choose x=10tan(t)? Is there an easy way to decide what you would replace x with?

Thank you for all of you guy's help.
Stephen

 

[Hitman2-2]

$$\int$$1/[(100sec(t)^2)^(3/2)] * 10sec(t)^2 dt
=
$$\int$$ [10sec(t)^2]/[100sec(t)^3] dt

not quite; remember to raise 100 to 3/2 power on the left hand side.

and if we solve this from -5 to 10
wouldn't it be (1/10)sin(10) - (1/10)sin(-5)

no

or
would it be (1/10)sin(tan(1)^(-1)) - (1/10)sin(tan(-.5)^(-1))
since x goes from -5 to 10, using x=10tan(t), would t=tan(10/10)^(-1) and t=tan(-5/10)^(-1)?

yes, except that instead of 1/10 you should have 1/100 (see above). In fact, you don't have to work that hard. Remember that we substituted $$x = 10 \tan (t)$$ so $$x/10 = \tan (t)$$.
Now what does this tell us? Draw a right triangle with one leg having length 10 and the other length x. Then the angle opposite side x is t and the hypotenuse is
$$\sqrt{100 + x^2}$$.
From this triangle we see that
$$\sin (t) = \frac{x}{\sqrt{100 + x^2}}$$.
So your antiderivative is
$$\frac{x}{100 \sqrt{100 + x^2}}$$
and you can work with the original limits of integration, x = -5 and x = 10.

And if so why did we choose x=10tan(t)? Is there an easy way to decide what you would replace x with?

Generally, when an integral has an odd power of $$\sqrt{x^2 + a^2}$$, the correct substitution is $$x = a \tan (\theta)$$ because for one thing, substituting this helps to get rid of the square root and for most elementary integrals, the integrand becomes some power of cosine or secant, which is more manageable.