Question

Integrate \[\int {\dfrac{{1 + \sin x}}{{\sin x\left( {1 + \cos x} \right)}}} \,dx\]?

Answer 1

Hint:In order to solve this question first, we assume a variable equal to the given integration. Then we make a substitution like \[t = \tan \dfrac{x}{2}\] and find the value of all other terms in terms of a new variable by using the formulas. Then simplify that expression and split all the parts and then integrate all the parts separately and again put the value in variable x.


Complete step by step answer:
Let \[I = \int {\dfrac{{1 + \sin x}}{{\sin x\left( {1 + \cos x} \right)}}} \,dx\]
To solve this integration we use a substitution method.
After substituting \[t = \tan \dfrac{x}{2}\] we find the value of \[\sin x\]and \[\cos x\] in terms of \[y\]
\[t = \tan \dfrac{x}{2}\]
On differentiating both sides with respect to \[x\].
\[dt = \dfrac{1}{2}{\sec ^2}\dfrac{x}{2}dx\]
Now converting sec trigonometry function in terms of tan trigonometry function by using the identity.
\[dt = \dfrac{1}{2}\left( {1 + {{\tan }^2}\dfrac{x}{2}} \right)dx\]
Now putting the value of \[\tan \dfrac{x}{2}\] in terms of \[t\].
\[\dfrac{{2dt}}{{1 + {\operatorname{t} ^2}}} = dx\]
Now using the formula of half angle in terms of tan trigonometry function.
\[\sin x = \dfrac{{2\tan \dfrac{x}{2}}}{{{{\tan }^2}\dfrac{x}{2} + 1}}\] and \[\cos x = \dfrac{{1 - {{\tan }^2}\dfrac{x}{2}}}{{1 + {{\tan }^2}\dfrac{x}{2}}}\]
Now putting the value of \[\tan \dfrac{x}{2}\] in terms of \[t\] in both the formulas.
\[\sin x = \dfrac{{2t}}{{{t^2} + 1}}\] and \[\cos x = \dfrac{{1 - {t^2}}}{{1 + {\operatorname{t} ^2}}}\]
Now putting all these values in the integration.
\[I = \int {\dfrac{{1 + \dfrac{{2t}}{{{t^2} + 1}}}}{{\dfrac{{2t}}{{{t^2} + 1}}\left( {1 + \dfrac{{1 - {t^2}}}{{1 + {\operatorname{t} ^2}}}} \right)}}} \,\dfrac{{2dt}}{{1 + {\operatorname{t} ^2}}}\]
Now on taking the LCM in numerator and denominator.
\[I = \int {\dfrac{{\dfrac{{1 + {\operatorname{t} ^2} + 2t}}{{{t^2} + 1}}}}{{\dfrac{{2t}}{{{t^2} + 1}}\left( {\dfrac{{1 + {\operatorname{t} ^2} + 1 - {t^2}}}{{1 + {\operatorname{t} ^2}}}} \right)}}} \,\dfrac{{2dt}}{{1 + {\operatorname{t} ^2}}}\]
On canceling the common terms.
\[I = \int {\dfrac{{1 + {\operatorname{t} ^2} + 2t}}{{2t}}} \,dt\]
Now splitting all these terms.
\[I = \int {\dfrac{1}{{2t}}dt + \int {\dfrac{{{t^2}}}{{2t}}dt + \int {\dfrac{{2t}}{{2t}}dt} } } \]
Now simplifying all these terms.
\[I = \dfrac{1}{2}\int {\dfrac{1}{t}dt + \dfrac{1}{2}\int {tdt + \int {dt} } } \]
Now integrating all the parts.
\[I = \dfrac{1}{2}\ln t + \dfrac{1}{2}\dfrac{{{t^2}}}{2} + t + c\]
Now again putting the value of \[t\] in terms of \[\tan \dfrac{x}{2}\]
\[I = \dfrac{1}{2}\ln \tan \dfrac{x}{2} + \dfrac{{{{\tan }^2}\dfrac{x}{2}}}{4} + \tan \dfrac{x}{2} + c\]
Here c is the constant of the integration.
The integration of \[\int {\dfrac{{1 + \sin x}}{{\sin x\left( {1 + \cos x} \right)}}} \,dx\] is-
\[I = \dfrac{1}{2}\ln \tan \dfrac{x}{2} + \dfrac{{{{\tan }^2}\dfrac{x}{2}}}{4} + \tan \dfrac{x}{2} + c\]

Note: In order to solve these types of questions students must have a knowledge of all the trigonometry identities and formulas and must have good practice to substitute the values. There are many places where students often make mistakes so take a look while solving the integration finding the values in terms of another variable.