Problem on integration!! need answer for this.. pl reply how to solve

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• Jul 19th 2007, 04:19 AM
kamaksh_ice
Problem on integration!! need answer for this.. pl reply how to solve
how to solve this?? ∫(cos4x-cos4α)/(cosx-cosα)dx. Also the limits are given to be 0 to Pi. Reply me the entire solution.. Its urgent pls
• Jul 19th 2007, 05:31 AM
topsquark
Quote:

Originally Posted by kamaksh_ice
how to solve this?? ∫(cos4x-cos4α)/(cosx-cosα)dx. Also the limits are given to be 0 to Pi. Reply me the entire solution.. Its urgent pls

Is this
$\displaystyle \int_0^{\pi} \frac{cos^4(x) - cos^4(a)}{cos(x) - cos(a)} dx$
or
$\displaystyle \int_0^{\pi} \frac{cos(4x) - cos(4a)}{cos(x) - cos(a)} dx$

-Dan
• Jul 19th 2007, 06:43 AM
kamaksh_ice
Answer me pls
It is
$\displaystyle \int_0^{\pi} \frac{cos(4x) - cos(4a)}{cos(x) - cos(a)} dx$
http://www.mathhelpforum.com/math-he...731f4a47-1.gif
• Jul 19th 2007, 10:57 AM
topsquark
Quote:

Originally Posted by kamaksh_ice
It is
$\displaystyle \int_0^{\pi} \frac{cos(4x) - cos(4a)}{cos(x) - cos(a)} dx$

This problem boils down to a factoring problem:

$\displaystyle cos(4x) = 8cos^4(x) - 8cos^2(x) + 1$
$\displaystyle cos(4a) = 8cos^4(a) - 8cos^2(a) + 1$

So
$\displaystyle cos(4x) - cos(4a) = 8[cos^4(x) - cos^4(a)] - [cos^2(x) - cos^2(a)]$

$\displaystyle = 8[cos^2(x) - cos^2(a)][cos^2(x) + cos^2(a)]- 8[cos^2(x) - cos^2(a)]$

$\displaystyle = 8(cos(x) - cos(a))[cos(x) + cos(a)][cos^2(x) + cos^2(a)]$ $\displaystyle - 8(cos(x) - cos(a))[cos(x) + cos(a)]$

So
$\displaystyle \int_0^{\pi} \frac{cos(4x) - cos(4a)}{cos(x) - cos(a)} dx$ $\displaystyle = \int_0^{\pi} (8[cos(x) + cos(a)][cos^2(x) + cos^2(a)] - 8[cos(x) + cos(a)]) dx$

$\displaystyle = 8 \int_0^{\pi}(cos^3(x) + cos(a) \cdot cos^2(x) - cos(x) + (cos^2(a) - cos(a))) dx$

all of which are now (relatively) elementary integrals.

-Dan