pls answer this question fast
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Let I = ∫
cos 2x − cos 2α
cos x − cos α
dx
=∫
(2 cos2x − 1) − (2 cos2α − 1)
(cos x − cos α)
dx
=∫
2(cos2x − 2 cos2α)(cos x − cos α)
dx
=2∫
(cos x − cos α)(cos x + cos α)
(cos x − cos α)
dx
=2∫(cos x + cos α) dx
=2 (sin x + x cos α) + C
So answer for root
√2(sin x+x cos α)+C
cos 2x − cos 2α
cos x − cos α
dx
=∫
(2 cos2x − 1) − (2 cos2α − 1)
(cos x − cos α)
dx
=∫
2(cos2x − 2 cos2α)(cos x − cos α)
dx
=2∫
(cos x − cos α)(cos x + cos α)
(cos x − cos α)
dx
=2∫(cos x + cos α) dx
=2 (sin x + x cos α) + C
So answer for root
√2(sin x+x cos α)+C
lemon3:
pls give full answer after 2 ∫(cosχ+cosα)dx
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