Question

Evaluate the integral \(\int {{{{\rm{(cosx + sinx)}}}^{\rm{2}}}} {\rm{cos2xdx}}\)

Short answer

The integral value of the given equation is\(\int {{{{\rm{(cosx + sinx)}}}^{\rm{2}}}} {\rm{cos2xdx = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{sin2x + }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ \times }}\left( {{\rm{ - }}\frac{{\rm{1}}}{{\rm{4}}}} \right){\rm{cos4x + C}}\).

Step-by-step solution

Expand the equation.

\(\begin{aligned}{c}\int {{{{\rm{(cosx + sinx)}}}^{\rm{2}}}} {\rm{cos2xdx = }}\int {\left( {{\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{ + 2sinxcosx + si}}{{\rm{n}}^{\rm{2}}}} \right)} {\rm{cos2xdx}}\\{\rm{ = }}\int {{\rm{(1 + 2sinxcosx)}}} {\rm{cos2xdx}}\end{aligned}\)

Known value \(\left( {{\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{ + si}}{{\rm{n}}^{\rm{2}}}{\rm{ = 1}}} \right)\)

\({\rm{ = }}\int {{\rm{(1 + sin2x)}}} {\rm{cos2xdx}}\) \(\;\;\;\;\;\;{\rm{(sin2x = 2sinxcosx)}}\)

Evaluate the equation.

Identification from a different aspect.

\(\begin{aligned}{c}{\rm{ = }}\int {{\rm{(cos2x + sin2xcos2x)}}} {\rm{dx}}\\{\rm{ = }}\int {\left( {{\rm{cos2x + }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{sin4x}}} \right)} {\rm{dx}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{sin2x + }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ \times }}\left( {{\rm{ - }}\frac{{\rm{1}}}{{\rm{4}}}} \right){\rm{cos4x + C}}\end{aligned}\)

Therefore, the integral value of the given equation is\(\int {{{{\rm{(cosx + sinx)}}}^{\rm{2}}}} {\rm{cos2xdx = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{sin2x + }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ \times }}\left( {{\rm{ - }}\frac{{\rm{1}}}{{\rm{4}}}} \right){\rm{cos4x + C}}\).