d) cos(x)/1+sin^2(x)
let u = sin(x)
du = cos(x)
int cos(x)/(1+u^2) {du/cos(x)}
cancel the cosine
int 1/(1+u^2) du
it becomes inverse tan integrate
tan^-1(u) + C
tan^-1(sin(x)) + C
Antiderive:
a) 3-x/1+9x^2 (split the fraction it says)
b) sin(-2x)/[1-cos(2x)]^5
c) -2/sqroot(9-x^2)
d) cox(x)/1+sin^2(x)
2. Consider that graph of y=-x^2+4x-3 and the area that lies under the curve in quad 1.
a)estimate the area using sumseq with 5 rectangles and using left endpoints
b)calculate exact area (is this fnInt function ?)
c)calc. average value of the function in the first quadrant.
3. Determine all relative max,min, inflection points (correct to 2 decimals)
for y=.2x^3-x^2-x^2-.8x+4. exhibit a labeled graph
Thanks for the help guys, this is part of my study guide for the final, i will finish it tomorrow and check my answer with the site to see what i did wrong. Thanks in advance!
b) sin(-2x)/[1-cos(2x)]^5
from trig identity
int (-2sin(x)cos(x)) / (1-(1-2sin^2(x)))^5 dx
-2 int (sin(x)cos(x)) / (2sin^2(x))^5 dx
cancel the sins
-2 int (cos(x)) / (32sin^9(x))
-2/32 int (cos(x)) / (u^9) {du / cos(x)} <<< derivative of sin
-1/16 int 1 / u^9 du
-1/16 [u^(-9+1)/(-8)]
substitute
1/128 * sin^-8(x)
1/128 * csc^8(x) + C
or
csc^8(x)/128 + C
a) (3-x)/(1+9x^2) (split the fraction it says)
int (3/1+9x^2) - int(x / (1+9x^2))
for integral 1 let u = 3x, du = 3dx
int (3/(1+u^2)) {du/3}
int (1/(1+(3x)^2)
tan^-1(3x)
for integral 2 let u = (1+9x^2), du = 18xdx
int (x/(1+(u)) {du /18x}
cancel tha x
1/18 int (1/(1+9x^2))
1/18 ln(1+9x^2)
tan^-1(3x) - 1/18 ln(1+9x^2) + C