Question

Find the general indefinite integral.

\(\int {\frac{{{\rm{sinx}}}}{{{\rm{1 - si}}{{\rm{n}}^{\rm{2}}}{\rm{x}}}}{\rm{dx}}} \)

Short answer

\(\int {\frac{{{\rm{sinx}}}}{{{\rm{1 - si}}{{\rm{n}}^{\rm{2}}}{\rm{x}}}}{\rm{dx}}} {\rm{ = secx + tanx + x + C}}\)

Step-by-step solution

Rationalising the equation.

\(\begin{array}{c}\int {\frac{{{\rm{sinx}}}}{{{\rm{1 - sinx}}}}\frac{{{\rm{(1 + sinx)}}}}{{{\rm{(1 + sinx)}}}}{\rm{dx}}} = \int {\frac{{{\rm{sinx + si}}{{\rm{n}}^{\rm{2}}}{\rm{x}}}}{{{\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{x}}}}{\rm{dx}}} \\\end{array}\)

Simplifying the equation

\(\begin{array}{c}{\rm{ = }}\int {\frac{{{\rm{sinx + si}}{{\rm{n}}^{\rm{2}}}{\rm{x}}}}{{{\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{x}}}}{\rm{dx}}} \\{\rm{ = }}\int {{\rm{secxtanxdx + }}\int {{\rm{ta}}{{\rm{n}}^{\rm{2}}}{\rm{x}}} } {\rm{dx}}\\{\rm{ = secx + }}\int {{\rm{(se}}{{\rm{c}}^{\rm{2}}}{\rm{x - 1)}}} {\rm{dx = sec + tanx + x + C}}\\\end{array}\)

Therefore,\(\int {\frac{{{\rm{sinx}}}}{{{\rm{1 - si}}{{\rm{n}}^{\rm{2}}}{\rm{x}}}}{\rm{dx}}} {\rm{ = secx + tanx + x + C}}\)