Question

Evaluate the integral

Answer 1

A

Answer 2

\[\int\limits_{0}^{\frac{ \pi }{ 2 }} \sin^7\Theta \cos^5\Theta dTheta \]

Answer 3

Okay so broke it down so it looks like the following

Answer 4

\[\int\limits_{0}^{\pi/2}\sin^6\Theta \sin \Theta \cos^4 \Theta \cos \Theta\]

Answer 5

Then I took it a step further

Answer 6

@iambatman

Answer 7

\[\int\limits_{0}^{\pi/2} (\sin^6 \Theta) (1-\cos \Theta) (\cos^4 \Theta) (1-\sin^2 \Theta)\]

Answer 8

But I am stuck from here, should I start u sub or are there more identities?

Answer 9

\[\int\limits \sin^7xcos^5x dx = \int\limits \sin^7xcos^4xcosxdx = \int\limits \sin^7x(1-\sin^2x)^2cosx dx\] now let u = sinx

Answer 10

remember to change the limits as well

Answer 11

okay, let me understand this, you only changed one of the odd powers to become even and not both

Answer 12

Yes

Answer 13

and it is (1-sin^2)^4 to accommodate the cos^4

Answer 14

^2 and yes

Answer 15

right okay, let me right this down real quick before I continue

Answer 16

That's the only tricky part, after it's a breeze :P

Answer 17

okay so I am going to have u = sinx and du = cosx correct?

Answer 18

Yup

Answer 19

du=cosxdx

Answer 20

You see it will get rid of all the cosx's now

Answer 21

And the sinx's will become u's

Answer 22

right so it looks like this \[\int\limits_{}^{} u^7 ( 1- u)^2 du\]

Answer 23

It should be (1-u^2)^2 but you get the idea

Answer 24

Then I should take care of the (1-u)^2 first? or plug in the bounds?

Answer 25

or distribute

Answer 26

\[u^7(1-u^2)^2 = u^7(1-u^2)(1-u^2)\]

Answer 27

Distribute

Answer 28

u^7 - u^4 (1-u^2)?

Answer 29

It's just algebra at this point you should get u^11-2u^9+u^7

Answer 30

Remember to change your limits, as I mentioned above as well.

Answer 31

right so I plug in the upper and lower limits into U to get the new bounds

Answer 32

Yeah, u = cosx, so at 0 and pi/2

Answer 33

I thought u = sinx

Answer 34

Yes, you're right! :P

Answer 35

sin of pi/2 is 1?

Answer 36

I was testing you *panics*

Answer 37

lol

Answer 38

Yes

Answer 39

okay so my new upper and lower limits are 1 and 0

Answer 40

where 1 is the upper limit and 0 is the lower limit

Answer 41

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

Answer 42

This is what your integral should look like at this step

Answer 43

okay so now I integrate

Answer 44

Yup :)

Answer 45

okay so I get \[\int\limits_{0}^{1}\frac{ 1 }{ 12 }u ^{12}- \frac{ 1 }{ 5 }u ^{10} + \frac{ 1 }{ 8 }u^8\]

Answer 46

You should not have the integral sign when you integrate, and yes

Answer 47

Doing the integration part*

Answer 48

okay so now I plug in the upper and lower limits

Answer 49

Yes

Answer 50

1/120?

Answer 51

Looks good to me!

Answer 52

awesome! thanks

Answer 53

No worries :)