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
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
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