Question

trig integral

Answer 1

\[\int\limits_{0}^{\pi/2} \sin^7 \Theta \cos^5 \Theta d \Theta\] and I made it into the following \[\int\limits_{0}^{\pi/2}(1-cosx^2)^5 \cos^{5}(x) \sin(x)dx \] and made u = \[\cos(x)\] however now I need to set the new bounds but I am getting \[\int\limits_{1}^{0}\] which is moving in the negative direction, would this be okay? or do I need to make cos the odd one out and put the bounds in terms of sin?

Answer 2

okay so I made cos the odd one out and got the following \[\int\limits_{0}^{1} u^7 (1-u^2)^3 du\]

Answer 3

Please note that keep in mind that your integrand function can be re-written as below:
\[{\left( {\sin \theta } \right)^7} \cdot {\left( {\cos \theta } \right)^5} = {\left( {\sin \theta } \right)^7} \cdot {\left( {1 - {{\left( {\sin \theta } \right)}^2}} \right)^2} \cdot \cos \theta \]

Answer 4

okay so I would still have the same bounds of 0 to 1

Answer 5

then \[u^7(1-u^2)^2 du\]

Answer 6

please set \[\sin \theta =u\]
so we have:
\[{\left( {\sin \theta } \right)^7} \cdot {\left( {\cos \theta } \right)^5}\;d\theta = {u^7} \cdot {\left( {1 - {u^2}} \right)^2} \cdot du\]

Answer 7

right, that's what I did

Answer 8

now do I need to do integration by parts?

Answer 9

no, please develop the expression:
\[{u^7} \cdot {\left( {1 - {u^2}} \right)^2} \]
as algebraic expression

Answer 10

you should get polynomial for u of 11th-grade

Answer 11

u^7 - 2u^9 + u^11

Answer 12

yes! Now it is easy to solve your integral for u

Answer 13

combine the U^7 and the u^11

Answer 14

no

Answer 15

no, since they are not similar monomial

Answer 16

\[\int\limits_{0}^{1}u^7 - 2u^9 + u^{11}\]

Answer 17

please use this:
\[\int {{u^n} = \frac{{{u^{n + 1}}}}{{n + 1}}} \]

Answer 18

you have to solve three ntegrals

Answer 19

oops...integrals

Answer 20

\[\frac{ 1 }{ 8 }U^8 - 5U^10 + \frac{ 1 }{ 12 }\]

Answer 21

that came out a little weird

Answer 22

1/12 u^12

Answer 23

ok!

Answer 24

please wait:
\[\frac{{{u^8}}}{8} + \frac{{{u^{12}}}}{{12}} - \frac{{{u^{10}}}}{5}\]

Answer 25

okay

Answer 26

1/120

Answer 27

that's right!

Answer 28

thanks!

Answer 29

thanks!