we can have three integration of 1 sinx cosx ( integration of 1 \/sinx+ cosx , integration of 1 \/sinx- cosx and integration of 1 \/sinx cosx). Here are formula of these integrals <\/p>\n\n\n\n
(I) integration of 1 \/sinx+ cosx<\/p>\n\n\n\n
$\\int \\frac {1}{sinx + cos x} \\; dx = \\frac {1}{\\sqrt{2}} \\log \\left| \\frac {\\left(\\tan\\left(\\frac{x}{2}\\right) – 1 + \\sqrt{2}\\right)}{\\left(\\tan\\left(\\frac{x}{2}\\right) -1- \\sqrt{2} \\right)}\\right| + C$<\/p>\n\n\n\n
or
$\\int \\frac {1}{sinx + cos x} \\; dx = \\frac {1}{\\sqrt{2}} \\ln |\\tan ( \\frac {x}{2} + \\frac {\\pi}{8}) | + C$<\/p>\n\n\n\n
(II) integration of 1 \/sinx- cosx<\/p>\n\n\n\n
$\\int \\frac {1}{sinx – cos x} \\; dx = \\frac {1}{\\sqrt 2} ln |\\frac {\\tan\\left(\\frac{x}{2}\\right) +1- \\sqrt 2}{\\tan\\left(\\frac{x}{2}\\right) + 2+ \\sqrt 2}| + C$<\/p>\n\n\n\n
or
$\\int \\frac {1}{sinx – cos x} \\; dx = \\frac {1}{\\sqrt{2}} \\ln |\\tan ( \\frac {x}{2} – \\frac {\\pi}{8}) | + C$<\/p>\n\n\n\n
(III) integration of 1 \/sinx cosx<\/p>\n\n\n\n
$\\int \\frac {1}{sin x cos x} \\; dx = \\frac{1}{2} \\ln |\\frac {(\\cos(2x) – 1)}{ (\\cos(2x) + 1)}| + C$<\/p>\n\n\n\n
or<\/p>\n\n\n\n
$\\int \\frac {1}{sin x cos x} \\; dx= \\ln |tan x| + C$<\/p>\n\n\n\n
The integral can be tricky due to the combination of sine and cosine functions in the denominator. A common approach is to use a tangent half-angle substitution, which simplifies the integration of trigonometric functions that involve both sine and cosine.<\/p>\n\n\n\n
Method 1<\/strong><\/p>\n\n\n\n The tangent half-angle substitution is $ t = \\tan\\left(\\frac{x}{2}\\right) $. This substitution leads to the identities:<\/p>\n\n\n\n $$ Substituting these into the integral, we get:<\/p>\n\n\n\n $$ $$ $$ $$ Now we know that Therefore<\/p>\n\n\n\n $$ where $ t = \\tan\\left(\\frac{x}{2}\\right) $ and $ C $ is the constant of integration.<\/p>\n\n\n\n To express this in terms of $ x $, we substitute back $ t = \\tan\\left(\\frac{x}{2}\\right) $, resulting in the final solution:<\/p>\n\n\n\n $$ Method 2<\/strong><\/p>\n\n\n\n $\\int \\frac {1}{sinx + cos x} \\; dx = \\int \\frac {1}{ \\sqrt 2 sin (x + \\frac {\\pi}{4})} \\; dx$ Now Therefore, original integral becomes<\/p>\n\n\n\n $=\\frac {1}{\\sqrt{2}} \\ln |\\tan ( \\frac {x}{2} + \\frac {\\pi}{8}) | + C$<\/p>\n\n\n\n The integral can be tricky due to the combination of sine and cosine functions in the denominator. A common approach is to use a tangent half-angle substitution, which simplifies the integration of trigonometric functions that involve both sine and cosine.<\/p>\n\n\n\n Method 1<\/strong><\/p>\n\n\n\n The tangent half-angle substitution is $ t = \\tan\\left(\\frac{x}{2}\\right) $. This substitution leads to the identities:<\/p>\n\n\n\n $$ Substituting these into the integral, we get:<\/p>\n\n\n\n $$ $$ $$ $$ $\\int \\frac {1}{x^2 – a^2} dx = \\frac {1}{2a} ln |\\frac {x-a}{x+a}| + C$ Therefore<\/p>\n\n\n\n $$ To express this in terms of $ x $, we substitute back $ t = \\tan\\left(\\frac{x}{2}\\right) $, resulting in the final solution:<\/p>\n\n\n\n $$ Method 2<\/strong><\/p>\n\n\n\n $\\int \\frac {1}{sinx – cos x} \\; dx = \\int \\frac {1}{ \\sqrt 2 sin (x – \\frac {\\pi}{4})} \\; dx$ Now Therefore, original integral becomes<\/p>\n\n\n\n $=\\frac {1}{\\sqrt{2}} \\ln |\\tan ( \\frac {x}{2} – \\frac {\\pi}{8}) | + C$<\/p>\n\n\n\n Method 1<\/p>\n\n\n\n $\\int \\frac {1}{sin x cos x} \\; dx = \\int \\frac {2}{ sin 2x} \\; dx $ Now Therefore, original integral becomes<\/p>\n\n\n\n $= \\ln |tan x| + C $<\/p>\n\n\n\n Method 2<\/p>\n\n\n\n We also know that<\/p>\n\n\n\n \\[ Therefore, original integral becomes<\/p>\n\n\n\n $= \\ln |\\frac {(\\cos(2x) – 1)}{ (\\cos(2x) + 1)}| + C $<\/p>\n\n\n\n I hope you find this article on integration of 1 sinx cosx useful and interesting. Please do provide the feedback<\/p>\n\n\n\n Other Integration Related Articles<\/p>\n\n\n\n
\\sin x = \\frac{2t}{1 + t^2}, \\quad \\cos x = \\frac{1 – t^2}{1 + t^2}, \\quad dx = \\frac{2}{1 + t^2} dt
$$<\/p>\n\n\n\n
\\int \\frac{1}{\\sin x + \\cos x} dx = \\int \\frac{1}{\\frac{2t}{1 + t^2} + \\frac{1 – t^2}{1 + t^2}} \\cdot \\frac{2}{1 + t^2} dt
$$<\/p>\n\n\n\n
=\\int \\frac{2}{2t + 1 – t^2} \\; dt
$$<\/p>\n\n\n\n
=\\int \\frac{2}{2 – (t^2 + 1 -2t)} \\; dt
$$<\/p>\n\n\n\n
=\\int \\frac{2}{(\\sqrt 2)^2 – (t-1)^2} \\; dt
$$<\/p>\n\n\n\n
$\\int \\frac {1}{a^2 – x^2} dx = \\frac {1}{2a} ln |\\frac {a+x}{a-x}| + C$
Proof<\/strong>
$\\frac {1}{a^2 – x^2} =\\frac {1}{2a}[ \\frac {1}{a-x} + \\frac {1}{a+x}]$
So
$\\int \\frac {1}{a^2 – x^2} dx $
$=\\frac {1}{2a}[ \\int \\frac {1}{a-x} dx + \\int \\frac {1}{x+a}]$
$= \\frac {1}{2a}[-ln |a-x| + ln |a+x| + C$
$=\\frac {1}{2a} ln |\\frac {a+x}{a-x}| + C$<\/p>\n\n\n\n
=\\frac {1}{\\sqrt{2}} \\left( \\log(t – 1 + \\sqrt{2}) – \\log(t – \\sqrt{2} – 1) \\right) + C
$$<\/p>\n\n\n\n
=\\frac {1}{\\sqrt{2}} \\log |\\frac {\\left(\\tan\\left(\\frac{x}{2}\\right) – 1 + \\sqrt{2}\\right)}{\\left(\\tan\\left(\\frac{x}{2}\\right) – \\sqrt{2} – 1\\right)}| + C
$$<\/p>\n\n\n\n
$=\\frac {1}{\\sqrt{2}} \\int \\csc (x + \\frac {\\pi}{4}) \\; dx$<\/p>\n\n\n\n
\\[
\\int \\csc(x) \\, dx = \\ln | \\tan \\frac {x}{2} | + C
\\]<\/p>\n\n\n\nProof of integration of 1\/sinx- cosx<\/h2>\n\n\n\n
\\sin x = \\frac{2t}{1 + t^2}, \\quad \\cos x = \\frac{1 – t^2}{1 + t^2}, \\quad dx = \\frac{2}{1 + t^2} dt
$$<\/p>\n\n\n\n
\\int \\frac{1}{\\sin x – \\cos x} dx = \\int \\frac{1}{\\frac{2t}{1 + t^2} – \\frac{1 – t^2}{1 + t^2}} \\cdot \\frac{2}{1 + t^2} dt
$$<\/p>\n\n\n\n
=\\int \\frac {2}{2t – 1 + t^2} \\; dt
$$<\/p>\n\n\n\n
=\\int \\frac {2} {(t^2 + 1 -2t) – 2} \\; dt
$$<\/p>\n\n\n\n
=\\int \\frac{2} {(t+1)^2 – (\\sqrt 2)^2} \\; dt
$$<\/p>\n\n\n\n
Proof<\/strong>
$\\frac {1}{x^2 – a^2} =\\frac {1}{2a}[ \\frac {1}{x-a} – \\frac {1}{x+a}]$
So
$\\int \\frac {1}{x^2 – a^2} dx $
$=\\frac {1}{2a}[ \\int \\frac {1}{x-a} dx – \\int \\frac {1}{x+a}]$
$= \\frac {1}{2a}[ln |x-a| – ln |x+a| + C$
$=\\frac {1}{2a} ln |\\frac {x-a}{x+a}| + C$<\/p>\n\n\n\n
=\\frac {1}{\\sqrt 2} ln |\\frac {t+1- \\sqrt 2}{t + 2+ \\sqrt 2}| + C
$$<\/p>\n\n\n\n
=\\frac {1}{\\sqrt 2} ln |\\frac {\\tan\\left(\\frac{x}{2}\\right) +1- \\sqrt 2}{\\tan\\left(\\frac{x}{2}\\right) + 2+ \\sqrt 2}| + C
$$<\/p>\n\n\n\n
$=\\frac {1}{\\sqrt{2}} \\int \\csc (x – \\frac {\\pi}{4}) \\; dx$<\/p>\n\n\n\n
\\[
\\int \\csc(x) \\, dx = \\ln | \\tan \\frac {x}{2} | + C
\\]<\/p>\n\n\n\nProof of integration of 1\/sinx cosx<\/h2>\n\n\n\n
$= 2 \\int \\csc 2x \\; dx$<\/p>\n\n\n\n
\\[
\\int \\csc(x) \\, dx = \\ln | \\tan \\frac {x}{2} | + C
\\]<\/p>\n\n\n\n
\\int \\csc(x) \\, dx = \\frac {1}{2} \\ln | \\frac {\\cos x -1}{\\cos x + 1} | + C
\\]<\/p>\n\n\n\n