Prove

∫ dx / (sec x + cosec x) = (1/2) (sinx-cosx) + 1/(2)^1/2 log | (tan x/2 - 1 - root2) / (tanx -1 + root2)|

Question

Prove &int; dx / (sec x + cosec x) = (1/2) (sinx-cosx) + 1/(2) 1/2 log | (tan x/2 - 1 - root2) / (tanx -1 + root2)|

Sumit Majumdar · Answer

Dear student, When we integrate: We get: This can be simplified to get the required result. Regards Sumit

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Prove ∫ dx / (sec x + cosec x) = (1/2) (sinx-cosx) + 1/(2) 1/2 log | (tan x/2 - 1 - root2) / (tanx -1 + root2)|

Prove

∫ dx / (sec x + cosec x) = (1/2) (sinx-cosx) + 1/(2)^1/2 log | (tan x/2 - 1 - root2) / (tanx -1 + root2)|

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Prove ∫ dx / (sec x + cosec x) = (1/2) (sinx-cosx) + 1/(2) 1/2 log | (tan x/2 - 1 - root2) / (tanx -1 + root2)|

Prove ∫ dx / (sec x + cosec x) = (1/2) (sinx-cosx) + 1/(2) 1/2 log | (tan x/2 - 1 - root2) / (tanx -1 + root2)|

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Prove

∫ dx / (sec x + cosec x) = (1/2) (sinx-cosx) + 1/(2)^1/2 log | (tan x/2 - 1 - root2) / (tanx -1 + root2)|