IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

hsbhatt

Evaluate $\frac {1}{\displaystyle \int _0^{\frac {\pi}{2}} \cos ^{2006}x \cdot \sin 2008 x\ dx}$

Indian_Dragon

my answer is coming 0.

here:

akhil_o

is it sin^2008 x or sin 2008 x?

hsbhatt

It is sin2008x. not to the power of.

sboosy

I_mn = ₍₀₎

^(pi/2)cos^mx(sin(nx))dx

which is

_{[0 ]}

^[pi/2]cos^mx d(-cos(nx) / n))

Using

Udv = UV-

Vdu

and substituing the limits

weget

I_mn = 1/n - m/n

cos ^m-1 x (sinx cos (nx)) dx

now

sin(n-1)x = sin nx cos x - cos nx sin x

substituting sinx cos nx from above eqn in the integral

we ger

I_mn = 1/n - m/n (I_mn) + m/n(I_{m-1 n-1})

so

I_mn= 1/(m+n) + m/(m+n) I_{m-1 n-1}

so I_mn = 1/(m+n) + m /(m+n)² + m²/(m+n)³ + ...general term (m^k/ (m+n)^k) I_{m-k n-k}

i think substituting appropriate values

the sum can be solved

hsbhatt

nope, sboosy, you dont get to cross 400 with this one.

sboosy

bhattji but is the procedure correct

kuldud

hey gud job indian dragon

its really amazin

well done

keep it up

hsbhatt

it does look like a possible JEE prob isnt it, i mean with that 2008 and all that.

elastiboysai

Hi sir

the prob does look scary !! bt finally turns out 2 b simple

_{[0 ]}

^{[pi/2 ]cos^2006x *sin(2007x+x)}

_{[0 ]}

^{[pi/2 ]sin2007xcos^2007x+cos2007xd(-cos^2007x/2007)}

^{apply parts for the second one and the first term cancels}

_{[0 ]}

^{[pi/2 ]sin2007xcos^2007x+[cos2007x*-cos^2007x/2007]-} _{[0 ]}

^{[pi/2 ]} ^{sin2007xcos^2007x}

^{so u get [cos2007x*-cos^2007x/2007}

^{between limits 0 and pi/.2}

^{whixh is 1/2007}

hsbhatt

DAV demolition squad at work again

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