How to integrate ∫(cos2x-cos2a)/(cosx-cosa)dx where a is constant
soumosree mallick , 7 Years ago
Grade 12
3 Answers
Arun
Last Activity: 7 Years ago
∫(cos 2x - cos 2a) / (cos x - cos a) dx = ∫((2cos^2 x - 1) - (2cos^2 a - 1)) / (cos x - cos a) dx = ∫(2 cos^2 x - 2 cos^2 a) / (cos x - cos a) dx = ∫[2(cos x - cos a)(cos x + cos a)] / (cos x - cos a) dx = ∫2(cos x + cos a) dx = 2 sin x + x cos a + c.
Rakshit Puri
Last Activity: 6 Years ago
Hsbs
∫(cos 2x - cos 2a) / (cos x - cos a) dx = ∫((2cos^2 x - 1) - (2cos^2 a - 1)) / (cos x - cos a) dx = ∫(2 cos^2 x - 2 cos^2 a) / (cos x - cos a) dx = ∫[2(cos x - cos a)(cos x + cos a)] / (cos x - cos a) dx = ∫2(cos x + cos a) dx = 2 sin x + 2x cos a + c
ankit singh
Last Activity: 4 Years ago
= 2 sin x + 2x cos a + c= ∫2(cos x + cos a) dx = ∫[2(cos x - cos a)(cos x + cos a)] / (cos x - cos a) dx = ∫(2 cos^2 x - 2 cos^2 a) / (cos x - cos a) dx ∫((2cos^2 x - 1) - (2cos^2 a - 1)) / (cos x - cos a) dx
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