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Harsh Garg 7 years, 11 months ago

(1+cosecx)tanx + log|cosecx-cotx| + c
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Tanmay Dixit Dixit 7 years, 11 months ago

Put tanx=t^2
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Sudhanshu Sharma 7 years, 11 months ago

Aproova ji. Cheating ni.... half se jyada kr leta hu.
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Sudhanshu Sharma 7 years, 11 months ago

Mind free hokr chap. Ko read kro. Plz
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Esha Modi 7 years, 11 months ago

Most probably March first week
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Pramod Kumar 7 years, 11 months ago

2cosxcosx-1
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Shambhu Kumar 7 years, 11 months ago

Pi/4
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Pramod Kumar 7 years, 11 months ago

1
'
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Ashish Verma 7 years, 11 months ago

Use voluntarily property

Ankit Verma 7 years, 11 months ago

3

Yuvraj Gupta 7 years, 11 months ago

Kya question hai....
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Satyendra Yadav 7 years, 11 months ago

IAI= -19 A= [2 1 1] X= X B=2 [1 3 -1] y 5 [3 1 -2 ]. z 6 adj A = [5 1 -8 ] [3 -7 1] [-8 1 5] A' = 1/-19 [-5 3 -4] [-1 -7 3] [-8 1 5 ] X= A'B 1/-19 [-5 3 -4] [2] [-1-7 3] [5] [-8 1 5] [6] = 1/-19 [-19] [-19] [ 19] :- X= 1 ; y= 1 ; z= -1 answer
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Anushka Banerjee 7 years, 11 months ago

Let 0≤x≤π/2 √(cos2x/cosx) =√(1-tan²x) ----- simple trig manipulation Let tanx = siny, -π/2≤y≤π/2 √(1-tan²x) = cosy, dx = cosy/(1+sin²y)dy, y = sin⁻¹(tanx) Thus, ∫√(cos2x/cosx)dx =∫cos²y/(1+sin²y) dy =∫-1 + 2/(cos²y+2sin²y)dy ---- simple trig manipulation = -y + ∫2/(cos²y+2sin²y)dy = -y + ∫2/(cos²y+2sin²y) * (sec²y/sec²y) dy ---- times sec²y/sec²y = -y + ∫2sec²y / (1 + 2tan²y) dy = -y + ∫√2 * (√2sec²y) / (1 + (√2tany)²) dy ---- manipulate a little = -y +√2 tan⁻(√2tany) + c ---- by u sub = - sin⁻¹(tanx) + √2 tan⁻¹(√2tan(sin⁻(tanx)) + c ----- take y back in = - sin⁻¹(tanx) + √2 tan⁻¹(√2tanx / √(1-tan²x)) + c Since √(cos2x/cosx) is even and - sin⁻¹(tanx) + √2 tan⁻¹(√2sinx / √(cos2x)) odd. The other "domain", π/2≤x≤ π, will lead to the same answer. The answer is confirmed by taking derivatives of - sin⁻¹(tanx) + √2 tan⁻¹(√2tanx / √(1-tan²x)) + c Some trig manipulations will result in the same solution as Wolfram|Alpha's: - sin⁻¹(tanx) = - tan¹(sinx/√cos2x) √2 tan⁻¹(√2tanx / √(1-tan²x)) = √2sin⁻¹(√2sinx)
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Madhavi Likhar 7 years, 11 months ago

Expand cos2x then divide separately

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