Trigonometric Integrals
Logarithmic and exponential Integrals
Definite Integrals with Chain Rule
Areas
Solids of rev
100

int (sin3(x)cos(x)dx)


sin4(x) / 4  +C

100

int ( e-8xdx)

- e-8x / 8 +C

100

int (x2-4x+6)dx from 1 to 8

259/3 

100

Find the area between y=8x-x2 and y=2x

A=36u2
100

Calculate de volume of the solid obtained by revolving the region under de graph f(x)=x+3  over the interval [0,9]

V=567pi u3

200

int(cot (33x) dx)

ln ( sin (33x)) / 33  +C

200

int (e-x6(-6x5) dx)

e-x6 +C

200

int ( dx /cubic root ((8+4x)2)) form 0 to 14

3 / 2

200

Find the area between y=16-x2 and y=x2-2

A=72u2

200

Calculate de volume of the solid obtained by revolving the region under de graph f(x)=sqrt (x-1)  over the interval [0,7]

V=18 pi u3

300

int (x31 sin (x32) dx)

- cos (x32) / 32 +C

300

int (ecos 27x sin(27x)dx)

-ecos 27x / 27  +C

300

int (x3(2x4+3)2dx)

49 / 12

300
Find the area between y=x2+2x+3 and y=2x+7
A=2 u2
300

Calculate de volume of the solid obtained by revolving the region under de graph f(x)=x+1  over the interval [0,2]

26 pi /3 u3

400

int (cos x / sin14 (x)  dx)

- 1 / 13 sin13 (x) +C

400

int (4-x dx)

- 4-x / ln (4)  +C

400

int ( 6 cos (pi(x)/2 dx) from 0 to 1

12 / pi

400

Find the area below y=16-x2 and the x-axis

A=256/3 u2

400

Calculate de volume of the solid obtained by revolving the region under de graph f(x)=2x+1  over the interval [0,3]

A=57 pi u3

500

int ( (sec (2x-56))2 dx)

-tan (2x-56) / 2 +C

500

int (x6 5x7 dx)

5x7 / 7ln (5) +C

500

int ( 1 / sqrt (3x +7) dx) from 3 to 6

2 / 3

500

Find the area below y=-4x+7 between [0,1]

A=5u2

500

Calculate de volume of the solid obtained by revolving the region under de graph f(x)=2sqrt (x+1)  over the interval [0,3]

V=18 pi u3