A couple example problems that when I see I know exactly what to do.
1. S 3/(x-13)^6 dx
*basic substitution
u=x-13 du= 1
substitute: 8 S 1/u^6 = 8 S u^-6
integrate: (1/5)(8) S u^-5
plug in: -8/5(x-13)^-5 +C
which can be written as -8/5(x-13)^5 +C
2. S sinxcos^4x dx
*basic substitution b/c cos and sin are direct derivatives of each other
u=cosx du= -sinx
substitute: -S u^4
integrate: -1/5u^5
plug in: (-1/5)cos^5x +C
rewritten: -cos^5x/5 +C
3. S cos^3 6x dx
*break it up
break: cos^26x(cos6x)
identity: (1-sin^26x)cos6x
multiply in: S cos 6x - cos6xsin^26x
*substitute for each
u=6x u=sin 6x
du=6 du=6cos6x
1/6 S cosu - 1/6 S u^2
3(sin6x/6) - sin^36x/18
3sin6x/18 - sin^36x/18
sin6x(3-sin^26x)/18
4. lim 28-7x+4x^2/5x^2 -7
x-->infinity
*you may think L"Hopital's Rule BUT it is as x-->infinity so use your limit rules!
exponents equal each other so divide coefficients
=4/5
5. infinity S 2 3/x^5 (diverge or converge?)
Let me know if this is right!
a S 2 3x^-5 = -3(1/4)x^-4
3/4 x^-4
lim x--> infinty 3/4x^4
= 0 +3/64
so it = 3/64?
NOW MY BIG QUESTIONS:
How do you do improper integrals if its bounds are like 9 to 11, like no infinity?
How do you know you can use synthetic division on an integral?
Anyone know some tricks about trig sub?
How you do chasing the rabbit again?
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you know to use synthetic division whenever your top exponent is greater than your bottom exponent.
ReplyDeleteexample:
S x^2/x+1
you would set the bottom equal to 0. solve for x, & then whatever you get goes in your box.
then you just follow through with synthetic division... :)