Question

Evaluate the integrals.

Answer 1

0 to (pi/4) cos x sin^3

Answer 2

Hint:

let u = sin(x)

du/dx = cos(x)

du = cos(x)*dx

Answer 3

\[\int\limits_{0}^{\pi/4}cosxsin^3x\] well, you can takeout a sin x resulting in \[\int\limits_{0}^{\pi/4}cosxsin^2xsinxdx\]

Answer 4

And then you can decide which one is easier to substitute using u-substitution. And as @jim_thompson5910 stated, use that u-sub to further solve your problem

Answer 5

okay

Answer 6

\[\int\limits_{0}^{\pi/4}cosx(1-\cos ^{2}x)sinxdx\]

Answer 7

okay

Answer 8

so what do i do next?

Answer 9

hint: use u-sub (see my post above) to get


\[\large \int\limits_{0}^{\pi/4}cosxsin^3x*dx\]


\[\large \int\limits_{0}^{\pi/4}sin^3x*cosx*dx\]


\[\large \int\limits_{0}^{\pi/4}u^3*du\]


Integrate with respect to u, then plug u = sin(x) back in and evaluate the limits

Answer 10

okay

Answer 11

so i'm integrating \[\int\limits_{0}^{\pi/4} u^3 dx\] ?

Answer 12

du not dx

Answer 13

times du not dx i mean

Answer 14

yes

Answer 15

okay thanks

Answer 16

np

Answer 17

remember to sub sin(x) back in for u after you've integrated

Answer 18

okay

Answer 19

so i'm sub sin(x) back into 0 to pi/4 u^3 du?

Answer 20

no you integrate u^3 du first to get

(1/4)*u^4 + C

THEN you sub sin(x) back in

Answer 21

Actually, no you don't have to substitute the sin back in because you've already changed the limits of integration

Answer 22

If you didn't change the limits then you would simply put in the values

Answer 23

\[\int\limits_{0}^{\frac{ \pi }{ 4 }}\sin^3cosxdx = \int\limits_{0}^{\frac{ \pi }{ 4 }}u^3du \] using what jim said, you chage your limits of integration by inputting your limits of integration back into your u-sub: so \[x=0, u=\sin0=0\]
\[x=\frac{ \pi }{ 4 } , u=\sin \frac{ \pi }{ 4 }=\frac{ \sqrt{2} }{ 2 }\] so with these new found limits, you evaluate the integral \[\int\limits_{0}^{\frac{ \sqrt{2} }{ 2 }}u^3du=\frac{ 1 }{ 4 }u^4\] from 0 -> rad 2/2 , no need for resubstitution.

Answer 24

okay so thats the answer?

Answer 25

no you still need to evaluate at the limits

Answer 26

evaluate (1/4)u^4 from 0 to rad 2/2

Answer 27

okay

Answer 28

ookay i evaluated and i got this long weird number :/

Answer 29

what did you get

Answer 30

i got 0.951261 :/

Answer 31

not sure how you got that, what did you type in

Answer 32

i typed evaluate 0 to rad 2/2 1/4 u^4

Answer 33

\[\int\limits_{0}^{\frac{ \sqrt{2} }{ 2 }}\frac{ 1 }{ 4 }u^4\] = \[\frac{ 1 }{ 4}[(\frac{ \sqrt{2} }{ 2 })^4 -(0)^4]\]

Answer 34

fyi: i represented (1/4)u^4 in integral form so you can see what i mean by evaluating it from0 -> rad 2/2. THIS DOES NOT MEAN INTEGRATE IT!!

Answer 35

ohh ok

Answer 36

i get 1/16

Answer 37

1/16 is correct

Answer 38

Thanks jim! :)

Answer 39

np