Evaluate the integral of 1/y² Cos (1/y) dy
Evaluate the integral
"\\displaystyle\\int\\dfrac{1}{y^2}\\cos \\left(\\frac{1}{y}\\right) dy"
Note that
"d\\left(\\dfrac{1}{y}\\right)=\\left(\\dfrac{1}{y}\\right)'dy=-\\dfrac{1}{y^2}dy"
Then
"\\displaystyle\\int\\dfrac{1}{y^2}\\cdot\\cos \\left(\\frac{1}{y}\\right) dy ="
"\\displaystyle\\int\\cos \\left(\\frac{1}{y}\\right)\\cdot\\dfrac{1}{y^2} dy ="
"-\\displaystyle\\int\\cos \\left(\\frac{1}{y}\\right)\\cdot\\left(-\\dfrac{1}{y^2}\\right) dy ="
"-\\displaystyle\\int\\cos \\left(\\frac{1}{y}\\right)d\\left(\\dfrac{1}{y}\\right) ="
"-\\sin \\left(\\dfrac{1}{y}\\right)+C"
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