Evaluate the ∫2sec² β bβ/tanβ
Evaluate the "\\int \\frac{2\\sec^2{\\beta}d\\beta}{\\tan{\\beta}}" .
Solution:
"\\int \\frac{2\\sec^2{\\beta}d\\beta}{\\tan{\\beta}}=\\int \\frac{2d\\beta}{\\tan{\\beta}\\cos^2{\\beta}}=\\int \\frac{2d\\tan{\\beta}}{\\tan{\\beta}}="
"2\\ln{|\\tan{\\beta}|}+C=\\ln{(\\tan^2{\\beta})}+C"
Answer:"\\int \\frac{2\\sec^2{\\beta}d\\beta}{\\tan{\\beta}}=\\ln{(\\tan^2{\\beta})}+C" .
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