Question

Beta function....
interesting problems!

Answer 1

\[\rm~A. \int\limits_{-\pi/6}^{\pi/6}(cosx+\sqrt3sinx)^{1/4}dx\\ B.\int\limits_{-\pi/2}^{\pi/2}(1+sinx)^2\cdot \cos^3xdx\\ C.\int\limits_{0}^{\pi}x\cdot \cos^6x\cdot \sin^5xdx\]

Answer 2

@surjithayer @Astrophysics @Zarkon @Loser66

Answer 3

for the first one i multiplied and divided by 2^1/4

Answer 4

after that i get sin(pi/6+x)

Answer 5

now the substitution :(

Answer 6

For A, let \(\cos x + \sqrt{3}\sin x = u\)

Answer 7

ohokay
ill try that

Answer 8

And using your idea :

\(\cos x + \sqrt{3}\sin x = u\)

\(\dfrac{1}{2}\cos x + \dfrac{\sqrt{3}}{2} \sin x = \dfrac{1}{2}u\)

\(\cos \dfrac{\pi }{3}\cos x + \sin\dfrac{\pi}{3}\sin x = \dfrac{1}{2}u\)

\(\cos(x-\dfrac{\pi}{3}) = \dfrac{1}{2}u\)

\(-\sin(x-\dfrac{\pi}{3})\,dx = \dfrac{1}{2}du \)

\(-\sqrt{ 1-(\frac{1}{2}u)^2} \,dx= \dfrac{1}{2}du\)

Answer 9

wow

Answer 10

last step

Answer 11

please wait a min
ill brb

Answer 12

back

Answer 13

last step i did not understand Sir

Answer 14

\[\sin x = \sqrt{1-\cos^2 x}\]

Answer 15

okay
so the limits will change correct

Answer 16

0 to root 3

Answer 17

correct?

Answer 18

hey u der?

Answer 19

for B) let u= 1+sin , then du = cos dx
sin= u-1
sin^2 =(u-1)^2
1-sin^2 =cos^2 =2u -u^2
combine all, you get u^2(2u-u^2) du
limits change to 0, 2
done

Answer 20

wait u can take sin only?

Answer 21

what do you mean?

Answer 22

u= 1+ sin then sin = u-1, why not?

Answer 23

wait sir please

Answer 24

\[\rm For~B~part: 1+sinx=u~\\~x=\sin^{-1}(u-1)~\\~dx=\frac{ 1 }{ \sqrt{1-(u-1)^2} }du~\\~for:x=-\pi/2-->u=0~\\~~~~~~~~~x=\pi/2-->u=2\]

Answer 25

Looks part A cannot be expressed in terms of beta function alone

Answer 26

:(

Answer 27

https://www.wolframalpha.com/input/?i=Integrate [%28cos+x+%2B+sqrt%283%29sin+x%29^%281%2F4%29,+{x,-pi%2F6,pi%2F6}]

Answer 28

what??? why?? 1+ sinx = u , then du = cos x dx.

Answer 29

You make it so complicated.

Answer 30

but then without taking the angle i havent used it ever

Answer 31

ok, I go steps
\(u= 1+ sinx \rightarrow sinx= u-1\), then \(du= cos x dx\)
\(cos^3(x) = cos^2(x) *cos (x)= (1-sin^2(x) )* cos (x)\)
this cos x dx = du, hence all we need is manipulate 1-sin^2(x)
from \(sin x= u-1\\sin^2(x)=(u-1)^2= u^2-2u+1\\1-sin^2(x) = 1-(u^2-2u+1)\\=2u-u^2\)

Answer 32

Now, combine all
\((1+sinx)^2cos^2(x)cosxdx= u^2(2u-u^2)du\)
for the limits, you just plug as you did above, (WITHOUT taking inverse)
u = 1+ sin x
if x =-pi/2, sin x =-1, then u = 1-1=0
if x = pi/2, sin x=1, then u =1+1 =2

Answer 33

i reached till the above step
now let me read this :3

Answer 34

oh okay
i got that @Loser66

Answer 35

now

Answer 36

no more. it is noon here, I miss my bed. hehehe

Answer 37

okay :)
thanks for your efforts Sir

Answer 38

close this post, post the problem you didn't get yet to a new one. NO one is patient to troll down all the long post to help you. Good luck