integration for 2011 aspirants

hehe ..

Nishant sir ... can't it be done in some other way like:

(1-x⁴)⁷ can be broken down to (1-x⁴)⁶(1-x⁴)

The expr. boils down to 29/4 * [ 1 - ∫x⁴(1-x⁴)⁶dx∫(1-x⁴)⁶dx ]

then (1-x⁴)⁶ can be binomially expanded in both numerator and denominator and integrated [ higher terms neglected. ]

The solution might come in a better form...

@arka wat bhaiya was telling was a hilarious stuff

and its pretty big if u go for binomial expansion
u may use reduction

By the way ... Sonne can u please tell me of any easier way with lesser steps?????

i didnt get wat u mean by neglecting higher powers here ? can u explain it again

http://latex.codecogs.com/gif.latex?\int_{0}^{1}\left%20(%201-x^4%20\right%20)^7\mathrm{dx}=x.(1-x^4)^7|_0^1+\int_{0}^{1}28x^4(1-x^4)^6\mathrm{dx}\\%20\int_{0}^{1}\left%20(%201-x^4%20\right%20)^7\mathrm{dx}=\int_{0}^{1}28(1-x^4-1)(1-x^4)^6)\mathrm{dx}\\%20\int_{0}^{1}\left%20(%201-x^4%20\right%20)^7\mathrm{dx}=\int_{0}^{1}28\left%20((1-x^4)^6)-(1-x^4)^7%20\right%20)\mathrm{dx}\\%20\frac{29\int_{0}^{1}(1-x^4)^7\mathrm{dx}}{4\int_{0}^{1}(1-x^4)^6\mathrm{dx}}=7

Hey ... what's the use of pinked post here???????

Nobody has given the answer yet!!!!!

Let I = ₀ ∫ ¹ ( 1 - x ⁴ ) ⁷ dx

J = ₀ ∫ ¹ ( 1 - x ⁴ ) ⁶ dx

Let Us Part - Integrate " I " .

I = [ ( x ) ( 1 - x ⁴ ) ⁷ ] ₀ ¹ - ₀ ∫ ¹ ( 1 - x ⁴ ) ⁶ { - 7 . 4 x ³ } ( x ) dx

= 0 - ₀ ∫ ¹ ( 1 - x ⁴ ) ⁶ ( 28 ) ( 1 - x ⁴ - 1 ) dx

= - ( 28 ) ₀ ∫ ¹ ( 1 - x ⁴ ) ⁷ dx + ( 28 ) ₀ ∫ ¹ ( 1 - x ⁴ ) ⁶ dx

= - 28 I + 28 J

So , I = - 28 I + 28 J .

Or , J = 2928 I

Given Integral K = 294 IJ = 284 = 7

P . S --- Calculation Mistake Corrected .

I think ans. 29/5. Thanks ... Ricky for the second method.

@arka :
sonne has posted answer in that hidden section [3]

I = \frac{29\int_{0}^{1}( 1- x^4)^7dx}{4\int_{0}^{1}(1-x^4)^6 dx}

\frac{1}{I} = \frac{4\int_{0}^{1}(1-x^4)^6 dx}{29\int_{0}^{1}( 1- x^4)^7dx}

\frac{29}{4I} = \frac{\int_{0}^{1}(1-x^4)^6 dx}{\int_{0}^{1}( 1- x^4)^7dx}

\int_{0}^{1}( 1- x^4)^7 . 1 dx \rightarrow x(1-x^4)^7 |_{0}^{1} - \int_{0}^{1}7 ( 1- x^4)^6( -4x^3)(x)

0 +28 \int_{0}^{1} ( 1- x^4)^6(x^4)

28 \int_{0}^{1} ( 1- x^4)^6(x^4)

\frac{29. 28}{4I } = \frac{\int_{0}^{1}(1-x^4)^6}{\int_{0}^{1}(1-x^4)^6)(x^4+1-1)}

\frac{29. 28}{4I } = \frac{\int_{0}^{1}(1-x^4)^6}{\int_{0}^{1}(1-x^4)^6 -\int_{0}^{1}(1-x^4)^7}

\frac{29. 28}{4I } = \frac{1}{\frac{\int_{0}^{1}(1-x^4)^6 -\int_{0}^{1}(1-x^4)^7}{\int_{0}^{1}(1-x^4)^6}}

\frac{29. 28}{4I } = \frac{1}{1 -\frac{\int_{0}^{1}(1-x^4)^7}{\int_{0}^{1}(1-x^4)^6}}

\frac{29. 28}{4I } = \frac{1}{1 -\frac{4I}{29}}

\frac{29. 28}{4I } = \frac{29}{29 -{4I}}

\frac{ 7}{I } = \frac{1}{29 -{4I}}

(29)(7)- 28 I = I

\not{29}.7 = \not{29}I

I = \boxed 7