Question

Verify by differentiation that the formula is correct.

\(\int {{\rm{co}}{{\rm{s}}^{\rm{2}}}} {\rm{xdx = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{x + }}\frac{{\rm{1}}}{{\rm{4}}}{\rm{sin2x + C}}\)

Short answer

Formula is Verified.

Step-by-step solution

Simplifying the formula.

The formula can be verified by showing that differentiating the value of indefinite integral gives back the function.

\(\begin{aligned}{c}\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\frac{{\rm{1}}}{{\rm{2}}}{\rm{x + }}\frac{{\rm{1}}}{{\rm{4}}}{\rm{sin2x + C}}} \right)\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ + }}\frac{{\rm{1}}}{{\rm{4}}}{\rm{cos2x}}{\rm{.2}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ + }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{cos2x}}...............{\rm{(1)}}\end{aligned}\)

Verifying the formula.

Using equation \({\rm{(1)}}\)

\(\begin{aligned}{c}{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{(1 + cos2x) = }}\frac{{\rm{1}}}{{\rm{2}}}\left( {{\rm{1 + co}}{{\rm{s}}^{\rm{2}}}{\rm{x - si}}{{\rm{n}}^{\rm{2}}}{\rm{x}}} \right)\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left( {{\rm{1 - si}}{{\rm{n}}^{\rm{2}}}{\rm{x + co}}{{\rm{s}}^{\rm{2}}}{\rm{x}}} \right){\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left( {{\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{x + co}}{{\rm{s}}^{\rm{2}}}{\rm{x}}} \right)\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left( {{\rm{2co}}{{\rm{s}}^{\rm{2}}}{\rm{x}}} \right){\rm{ = co}}{{\rm{s}}^{\rm{2}}}{\rm{x}}\end{aligned}\)