Indistinct
Indistinct Indistinct
  • 17-12-2018
  • Mathematics
contestada

Solve the integral

I've come across some difficulties with this one..

Solve the integralIve come across some difficulties with this one class=

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LammettHash
LammettHash LammettHash
  • 17-12-2018

Multiply the integrand's numerator and denominator by [tex]1-\sin x[/tex], then simplify:

[tex]\displaystyle\int\frac{\mathrm dx}{1+\sin x}=\int\frac{1-\sin x}{1-\sin^2x}\,\mathrm dx=\int\frac{1-\sin x}{\cos^2x}\,\mathrm dx=\int(\sec^2x-\sec x\tan x)\,\mathrm dx[/tex]

You should be familiar with

[tex]\dfrac{\mathrm d}{\mathrm dx}\tan x=\sec^2x[/tex]

[tex]\dfrac{\mathrm d}{\mathrm dx}\sec x=\sec x\tan x[/tex]

so that

[tex]\displaystyle\int\frac{\mathrm dx}{1+\sin x}=\tan x-\sec x+C[/tex]

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