# U-Substitution Integrals

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• January 7th 2010, 02:34 PM
ali2141
U-Substitution Integrals
Hey guys, how would you do these? For all of these I can't figure out a way to cancel out the x's after I find u and du.

(x+1)^3/(x^2)dx

(x+1)/(x^2) dx

1/(1+4(x^2)) dx

x(x-10)^10 dx

(cosx)/(1+(sinx)^2) dx

(1+tanx)^3/(cosx)^2

Thanks :)
• January 7th 2010, 02:40 PM
Drexel28
Quote:

Originally Posted by ali2141
Hey guys, how would you do these? For all of these I can't figure out a way to cancel out the x's after I find u and du.

(x+1)^3/(x^2)dx

(x+1)/(x^2) dx

1/(1+4(x^2)) dx

x(x-10)^10 dx

(cosx)/(1+(sinx)^2) dx

(1+tanx)^3/(cosx)^2

Thanks :)

I have no idea what cancellation you are referring to.
• January 7th 2010, 02:42 PM
pickslides
Quote:

Originally Posted by ali2141

(x+1)/(x^2) dx

You may not need a substitution for this one.

$
\int \frac{x+1}{x^2} ~dx = \int \frac{x}{x^2} +\frac{1}{x^2}~dx = \int \frac{1}{x} +\frac{1}{x^2}~dx
$
• January 7th 2010, 02:43 PM
Quote:

Originally Posted by ali2141
Hey guys, how would you do these? For all of these I can't figure out a way to cancel out the x's after I find u and du.

(x+1)^3/(x^2)dx

(x+1)/(x^2) dx

1/(1+4(x^2)) dx

x(x-10)^10 dx

(cosx)/(1+(sinx)^2) dx

(1+tanx)^3/(cosx)^2

Thanks :)

It's not clear what you did here. If you want to integrate $\int\frac{(x+1)^3}{x^2}dx$, you can just expand the numerator $\int\frac{x^3+3x^2+3x+1}{x^2}dx$ and then split it up into a sum. Then you should be able to apply the general formula for polynomial integrands.
• January 7th 2010, 02:56 PM
ali2141
I was trying to use u-substitution.

Wait, so you mean do: x^3/x^2 + 3x^2/x^2 + 3x/x^2 + 1/x^2 ?
What would be the answer to that then?
• January 7th 2010, 03:12 PM
Quote:

Originally Posted by ali2141
I was trying to use u-substitution.

Wait, so you mean do: x^3/x^2 + 3x^2/x^2 + 3x/x^2 + 1/x^2 ?
What would be the answer to that then?

You don't see how to integrate that? If you're studying the substitution rule, I assume you must be familiar with the basic methods of integral calculus. Just use the sum rule, and integrate each term using the formula $\int ax^ndx=\frac{a}{n+1}x^{n+1}+C$
• January 7th 2010, 03:20 PM
ali2141
No, I get that if you have x^3, the antiderivative would be x^4/4
But I don't know how that would apply to something like x^3/x^2
• January 7th 2010, 03:24 PM
pickslides
Quote:

Originally Posted by ali2141
But I don't know how that would apply to something like x^3/x^2

$
\frac{x^3}{x^2} = \frac{x\times x \times x}{x \times x} = x
$
• January 7th 2010, 03:26 PM
ali2141
okay, i feel like an idiot now.
• January 7th 2010, 04:22 PM
TWiX
Quote:

You don't see how to integrate that? If you're studying the substitution rule, I assume you must be familiar with the basic methods of integral calculus. Just use the sum rule, and integrate each term using the formula $\int ax^ndx=\frac{a}{n+1}x^{n+1}+C$

n dont equal -1
• January 7th 2010, 04:39 PM
Prove It
Quote:

Originally Posted by ali2141
Hey guys, how would you do these? For all of these I can't figure out a way to cancel out the x's after I find u and du.

(x+1)^3/(x^2)dx

(x+1)/(x^2) dx

1/(1+4(x^2)) dx

x(x-10)^10 dx

(cosx)/(1+(sinx)^2) dx

(1+tanx)^3/(cosx)^2

Thanks :)

The first two have been taken care of.

for $\int{\frac{1}{1 + 4x^2}\,dx}$ you do not use a $u$ substitution. You use a trigonometric substitution.

$\int{\frac{1}{1 + 4x^2}\,dx} = \int{\frac{1}{4\left(\frac{1}{4} + x^2\right)}\,dx}$

$= \frac{1}{4}\int{\frac{1}{\frac{1}{4} + x^2}\,dx}$.

Now use the substitution $x = \frac{1}{2}\tan{\theta}$.

Note that $\frac{dx}{d\theta} = \frac{1}{2}\sec^2{\theta}$.

So $dx = \frac{1}{2}\sec^2{\theta}\,d\theta$.

Therefore

$= \frac{1}{4}\int{\frac{1}{\frac{1}{4} + x^2}\,dx} = \frac{1}{4}\int{\frac{1}{\frac{1}{4} + \left(\frac{1}{2}\tan{\theta}\right)^2}\,\frac{1}{ 2}\sec^2{\theta}\,d\theta}$

$= \frac{1}{8}\int{\frac{1}{\frac{1}{4}(1 + \tan^2{\theta})}\,\sec^2{\theta}\,d\theta}$

$= \frac{1}{2}\int{\frac{1}{1 + \tan^2{\theta}}\,\sec^2{\theta}\,d\theta}$

$= \frac{1}{2}\int{\frac{1}{\sec^2{\theta}}\,\sec^2{\ theta}\,d\theta}$

$= \frac{1}{2}\int{1\,d\theta}$

$= \frac{1}{2}\theta + C$.

Now remembering that $x = \frac{1}{2}\tan{\theta}$

$2x = \tan{\theta}$

$\theta = \arctan{2x}$.

So finally our answer is $\frac{1}{2}\arctan{2x} + C$.
• January 7th 2010, 04:46 PM
TWiX
Quote:

Originally Posted by ali2141
(cosx)/(1+(sinx)^2) dx

substitute u = sinx
then you will face a famous integral.
• January 7th 2010, 04:48 PM
Prove It
Quote:

Originally Posted by ali2141
Hey guys, how would you do these? For all of these I can't figure out a way to cancel out the x's after I find u and du.

(x+1)^3/(x^2)dx

(x+1)/(x^2) dx

1/(1+4(x^2)) dx

x(x-10)^10 dx

(cosx)/(1+(sinx)^2) dx

(1+tanx)^3/(cosx)^2

Thanks :)

For $\int{x(x - 10)^{10}\,dx}$ you could expand everything, but that could get messy.

Instead, integration by parts will need to be used.

Let $u = x$ so that $du = 1$.

Let $dv = (x - 10)^{10}$ so that $v = \int{(x - 10)^{10}\,dx} = \frac{1}{11}(x - 10)^{11}$.

Therefore $\int{x(x - 10)^{10}\,dx} = \frac{1}{11}x(x - 10)^{11} - \int{\frac{1}{11}(x - 10)^{11}\,dx}$

$= \frac{1}{11}x(x - 10)^{11} - \frac{1}{132}(x - 10)^{12} + C$

$= \frac{1}{11}(x - 10)^{11}\left[x - \frac{1}{12}(x - 10)\right] + C$

$= \frac{1}{11}(x - 10)^{11}\left(x - \frac{1}{12}x + \frac{5}{6}\right)$

$= \frac{1}{11}(x - 10)^{11}\left(\frac{11}{12}x + \frac{5}{6}\right)$

$= \frac{1}{132}(x - 10)^{11}(11x + 10)$.
• January 7th 2010, 04:51 PM
Prove It
Quote:

Originally Posted by ali2141
Hey guys, how would you do these? For all of these I can't figure out a way to cancel out the x's after I find u and du.

(x+1)^3/(x^2)dx

(x+1)/(x^2) dx

1/(1+4(x^2)) dx

x(x-10)^10 dx

(cosx)/(1+(sinx)^2) dx

(1+tanx)^3/(cosx)^2

Thanks :)

For $\int{\frac{(1 + \tan{x})^3}{\cos^2{x}}\,dx}$, note that

$\frac{1}{\cos^2{x}} = \sec^2{x}$.

So $\int{\frac{(1 + \tan{x})^3}{\cos^2{x}}\,dx} = \int{(1 + \tan{x})\sec^2{x}\,dx}$

Now let $u = 1 + \tan{x}$ so that $\frac{du}{dx} = \sec^2{x}$.

The integral becomes

$\int{u\,\frac{du}{dx}\,dx}$

$= \int{u\,du}$

$= \frac{1}{2}u^2 + C$

$= \frac{1}{2}(1 + \tan{x})^2 + C$.
• January 8th 2010, 10:10 AM
roshanhero
I am sorry but shouldnot there have been
$

\int{\frac{(1 + \tan{x})^3}{\cos^2{x}}\,dx} = \int{(1 + \tan{x})^3\sec^2{x}\,dx}
$
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