Evaluate the integral of sec 5θ tan 5θ dθ
Let us evaluate the integral of "\\int \\sec 5\u03b8 \\tan 5\u03b8 d\u03b8:"
"\\int \\sec 5\u03b8 \\tan 5\u03b8 d\u03b8=\\int \\frac{1}{\\cos 5\u03b8}\\frac{ \\sin 5\u03b8}{\\cos 5\u03b8} d\u03b8=\n-\\frac{1}{5}\\int \\frac{1}{\\cos^2 5\u03b8} d(\\cos 5\u03b8)=\n-\\frac{1}{5}\\int \\cos^{-2} 5\u03b8 d(\\cos 5\u03b8)=\n\n\n\\frac{1}{5}\\ \\cos^{-1} 5\u03b8 + C=\\frac{1}{5\\cos 5\u03b8}+C"
Comments
Leave a comment