# Integral problem

• Apr 4th 2009, 11:01 AM
LaraSoft
Integral problem
Hello!
Can someone help me with this problem (advanced section):

Calculate the indefinite integral $\int[(1+\sin(x))/(1-\sin(x))^2]dx$, not using a universal substitution $\tan(x/2)=t$ .

• Apr 4th 2009, 11:31 AM
Jester
Quote:

Originally Posted by LaraSoft
Hello!
Can someone help me with this problem (advanced section):

Calculate the indefinite integral $\int[(1+\sin(x))/(1-\sin(x))^2]dx$, not using a universal substitution $\tan(x/2)=t$ .

Multiply numerator and denominator by $(1 + \sin x)^2$ and use identities on the denominator to obtain $\cos^4x$ expand and integrate term by term.
• Apr 4th 2009, 12:50 PM
DeMath
Or so

$I = \int {\frac{{1 + \sin x}}{{{{\left( {1 - \sin x} \right)}^2}}}dx} = \int {\frac{{\left( {1 - \sin x} \right)\left( {1 + \sin x} \right)}}{{\left( {1 - \sin x} \right){{\left( {1 - \sin x} \right)}^2}}}} = \int {\frac{{{{\cos }^2}x}}{{{{\left( {1 - \sin x} \right)}^3}}}dx} =$

$= \frac{1}{2}\int {\cos xd\left( {{{\left( {1 - \sin x} \right)}^{ - 2}}} \right)} = \frac{{\cos x}}{{2{{\left( {1 - \sin x} \right)}^2}}} + \frac{1}{2}\int {\frac{{\sin x}}{{{{\left( {1 - \sin x} \right)}^2}}}dx} =$

$= \frac{{\cos x}}{{2{{\left( {1 - \sin x} \right)}^2}}} + \frac{1}{4}\int {\frac{{1 + \sin x - \left( {1 - \sin x} \right)}}{{{{\left( {1 - \sin x} \right)}^2}}}dx} =$

$= \frac{{\cos x}}{{2{{\left( {1 - \sin x} \right)}^2}}} + \frac{1}
{4}\underbrace {\int {\frac{{1 + \sin x}}{{{{\left( {1 - \sin x} \right)}^2}}}dx} }_I - \frac{1}{4}\int {\frac{{dx}}{{1 - \sin x}}} \Leftrightarrow$

$\Leftrightarrow I - \frac{1}{4}I = \frac{{\cos x}}{{2{{\left( {1 - \sin x} \right)}^2}}} - \frac{1}{4}\int {\frac{{dx}}{{1 - \sin x}}} \Leftrightarrow$

$\Leftrightarrow I = \frac{{2\cos x}}{{3{{\left( {1 - \sin x} \right)}^2}}} - \frac{1}{3}\int {\frac{{dx}}{{1 - \sin x}}}.$

$\int {\frac{{dx}}{{1 - \sin x}}} = \int {\frac{{1 + \sin x}}{{1 - {{\sin }^2}x}}dx} = \int {\frac{{1 + \sin x}}{{{{\cos }^2}x}}dx} = \int {\frac{{dx}}{{{{\cos }^2}x}}} + \int {\frac{{\sin x}}{{{{\cos }^2}x}}dx} =$

$= \tan x - \int {\frac{{d\left( {\cos x} \right)}}{{{{\cos }^2}x}}} = \tan x + \frac{1}{{\cos x}} + C = \tan x + \sec x + C.$

$I = \frac{{2\cos x}}{{3{{\left( {1 - \sin x} \right)}^2}}} - \frac{1}{3}\left( {\tan x + \sec x} \right) + C.$