U Substitution

• December 7th 2009, 01:19 AM
Lupus7874
U Substitution
i'm confused on which item should be my u and which should be my du, i've tried a few values on all problems but I feel my answer is wrong and that I'm overlooking something, if you could please help I would be eternally grateful!

Use U substitution to solve the following:
a)anti derivative of cos(x)/sin^2(x) dx top valuei/2 bottom value: pi/4
b) (4(tan^-1(x))^2)/(1+x^2) dx top value: 1 bottom value: -1
c) 5x(3x^2 - 12)^9 dx top value: 4 bottom value: 0
• December 7th 2009, 01:26 AM
mr fantastic
Quote:

Originally Posted by Lupus7874
i'm confused on which item should be my u and which should be my du, i've tried a few values on all problems but I feel my answer is wrong and that I'm overlooking something, if you could please help I would be eternally grateful!

Use U substitution to solve the following:
a)anti derivative of cos(x)/sin^2(x) dx top valuei/2 bottom value: pi/4
b) (4(tan^-1(x))^2)/(1+x^2) dx top value: 1 bottom value: -1
c) 5x(3x^2 - 12)^9 dx top value: 4 bottom value: 0

a) Substitute $u = \sin x$.
b) Substitute $u = \tan^{-1} x$.
c) Substitute $u = 3x^2 - 12$.

If you need more help, please post all your work and state exactly where you're stuck.
• December 7th 2009, 06:08 AM
Lupus7874
Ok for example a) I solve to the point of du*u^2 in said integral, but I'm just lost as to what to do when calculating the antiderivative of a product, surely I can't just apply the power rule in reverse can I?
• December 7th 2009, 07:22 AM
Quote:

Originally Posted by Lupus7874
Ok for example a) I solve to the point of du*u^2 in said integral, but I'm just lost as to what to do when calculating the antiderivative of a product, surely I can't just apply the power rule in reverse can I?

$\int_{\frac{\pi}{4}}^{\pi}\frac{cos(x)}{sin^2(x)}d x$

If $u=sin(x)$ then $du=cos(x)dx$ so $dx=\frac{du}{cos(x)}$. Substitute these values into the original integral to obtain $\int_{u_1}^{u_2}\frac{1}{u^2}du$. The limits of the integral are determined by the values of $u$ at the corresponding values of $x$.

This isn't a product. This can be evaluated using the basic formula:

$\int ax^ndx=\frac{a}{n+1}x^{n+1}+C$
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Secondly, there isn't a product rule for integration. There's only a sum and diffference rule. For product and quotients you have to deal with on a case-by-case basis. Sometimes you need to use either the substitution rule, integration by parts, partial fractions, etc. For example:

$\int sec^3(x)tan(x)dx$

We know that $\frac{d}{dx}sec(x)=sec(x)tan(x)$, so if we use $u=sec(x)$ then $du=sec(x)tan(x)dx$. This means that we can substitute $dx=\frac{du}{sec(x)tan(x)}$ into the original equation, canceling out the $tan(x)$ and leaving only $sec^2(x)$ which is equal to $u^2$.

Now we have $\int u^2du=\frac{1}{3}u^3+C=\frac{1}{3}sec^3(x)+C$

Therefore:

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