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JEE Revision Notes

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Illustration 1: Integrate the curve x/ (1+x4). (1978)
Solution:Let I = ∫xdx / (1 + x4)
= 1/2.∫ 2x / (1 + (x2)2) dx
Put x2 = u then 2xdx = du
Hence, I = ∫ du/ 2(1 + u2)
= 1/2 tan-1 u +c
= 1/2 tan-1(x2)+c

Illustration 2: Integrate sin x. sin 2x.sin 3x + sec2x. cos2 2x + sin4 x cos4 x. (1979)
Solution:Let I1 = ∫sin x. sin 2x. sin 3x dx
= 1/4 ∫ sin 4x + sin 2x - sin 6x)dx
= -cos 4x/16 -cos 2x/8 + cos 6x/24
I2 = ∫sec2x. cos2 2x dx
= ∫ sec2x (2cos2x – 1)2 dx
= ∫ (4cos2 x + sec2x- 4)dx
=∫(2cos 2x + sec2x-2)dx
= sin 2x + tan x – 2x
And I3 = ∫sin4x. cos4 x dx
= 1/128 ∫(3 - 4cos 4x + cos 8x)dx
= 3x/128 – sin 4x /128 + sin 8x/1024
Hence, I = I+ I2 + I3
= -cos 4x/16 -cos 2x/8 + cos 6x/24 + sin 2x + tan x – 2x + 3x/128 – sin 4x /128 + sin 8x/1024.

Illustration 3: For any natural umber m, evaluate
∫(x3m + x2m + xm) (2x2m + 3xm + 6)1/mdx, x > 0.(2002)
Solution: For any natural number m, the given integral can be written as
I = ∫(x3m + x2m + xm) [(2x3m + 3x2m + 6xm)1/m dx] / x
Hence I = ∫(2x3m + 3x2m + 6xm)1/m (x3m-1 + x2m-1 + xm-1)dx
Put (2x3m + 3x2m + 6xm) = t
Then 6x3m-1 + 6mx2m-1 + 6mxm-1)dx = dt
Hence I = ∫t1/m dt/6m
= [(2x3m + 3x2m + 6xm)(m+1)/m] / 6(m+1) + c

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