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# Notes | EduRev

## JEE : Notes | EduRev

The document Notes | EduRev is a part of the JEE Course JEE Revision Notes.
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Illustration 1: Integrate the curve x/ (1+x4). (1978)
Solution:Let I = âˆ«xdx / (1 + x4)
= 1/2.âˆ« 2x / (1 + (x2)2) dx
Put x2 = u then 2xdx = du
Hence, I = âˆ« 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 = âˆ«sin x. sin 2x. sin 3x dx
= 1/4 âˆ« sin 4x + sin 2x - sin 6x)dx
= -cos 4x/16 -cos 2x/8 + cos 6x/24
I2 = âˆ«sec2x. cos2 2x dx
= âˆ« sec2x (2cos2x â€“ 1)2 dx
= âˆ« (4cos2 x + sec2x- 4)dx
=âˆ«(2cos 2x + sec2x-2)dx
= sin 2x + tan x â€“ 2x
And I3 = âˆ«sin4x. cos4 x dx
= 1/128 âˆ«(3 - 4cos 4x + cos 8x)dx
= 3x/128 â€“ sin 4x /128 + sin 8x/1024
Hence, I = I+ I2 + I3
= -cos 4x/16 -cos 2x/8 + cos 6x/24 + sin 2x + tan x â€“ 2x + 3x/128 â€“ sin 4x /128 + sin 8x/1024.

Illustration 3: For any natural umber m, evaluate
âˆ«(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 = âˆ«(x3m + x2m + xm) [(2x3m + 3x2m + 6xm)1/m dx] / x
Hence I = âˆ«(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 = âˆ«t1/m dt/6m
= [(2x3m + 3x2m + 6xm)(m+1)/m] / 6(m+1) + c

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