How to integrate ∫(cos2x-cos2a)/(cosx-cosa)dx where a is constant
How to integrate ∫(cos2x-cos2a)/(cosx-cosa)dx where a is constant
soumosree mallick , 8 Years ago
Grade 12
Integral Calculus> How to integrate ∫(cos2x-cos2a)/(cosx-cos...
3 Answers
Arun
∫(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.
= ∫((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
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
= ∫((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
= 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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