Question

Evaluate the indefinite integral\(\int {\frac{{{\rm{sinx}}}}{{{\rm{1 + co}}{{\rm{s}}^{\rm{2}}}{\rm{x}}}}} {\rm{dx}}\).

Short answer

The indefinite integral value of the given equation is\(\int {\frac{{{\rm{sinx}}}}{{{\rm{1 + co}}{{\rm{s}}^{\rm{2}}}{\rm{x}}}}} {\rm{dx = - ta}}{{\rm{n}}^{{\rm{ - 1}}}}\left( {{\rm{cosx}}} \right){\rm{ + c}}\).

Step-by-step solution

Expand the number of variables in the equation.

Given: \({\rm{I = }}\int {\frac{{{\rm{sinx}}}}{{{\rm{1 + co}}{{\rm{s}}^{\rm{2}}}{\rm{x}}}}} {\rm{dx}}\)

Let \({\rm{u = cosx}} \to {\rm{du = ( - sinx)dx}} \to {\rm{dx = - }}\frac{{{\rm{du}}}}{{{\rm{sinx}}}}\)

Substitute \({\rm{x}}\)and \({\rm{dx}}\)for \({\rm{u}}\)and \({\rm{du}}\)

\({\rm{I = - }}\int {\frac{{\rm{1}}}{{{{\rm{u}}^{\rm{2}}}{\rm{ + 1}}}}} {\rm{du}}\)

Evaluate the equation.

Integrate \(\;\;{\rm{I = - ta}}{{\rm{n}}^{{\rm{ - 1}}}}{\rm{u + c}}\)

Return to the original position \(\;{\rm{u = cos}}\)

\({\rm{I = - ta}}{{\rm{n}}^{{\rm{ - 1}}}}\left( {{\rm{cosx}}} \right){\rm{ + c}}\)

Therefore, the indefinite integral value of the given equation is\(\int {\frac{{{\rm{sinx}}}}{{{\rm{1 + co}}{{\rm{s}}^{\rm{2}}}{\rm{x}}}}} {\rm{dx = - ta}}{{\rm{n}}^{{\rm{ - 1}}}}\left( {{\rm{cosx}}} \right){\rm{ + c}}\).