Question

Evaluate the integral\(\int {{\rm{cs}}{{\rm{c}}^{\rm{4}}}{\rm{4xdx}}} \).

Short answer

The integral value of the given equation is\(\int {{\rm{cs}}{{\rm{c}}^{\rm{4}}}{\rm{4xdx}}} {\rm{ = - }}\frac{{\rm{1}}}{{\rm{4}}}{\rm{cot(4x) - }}\frac{{\rm{1}}}{{{\rm{12}}}}{\rm{co}}{{\rm{t}}^{\rm{3}}}{\rm{(4x) + C}}\).

Step-by-step solution

Expand the equation.

Known value

\({\rm{cs}}{{\rm{c}}^{\rm{2}}}{\rm{x = 1 + co}}{{\rm{t}}^{\rm{2}}}{\rm{x}}\)

So,

\(\begin{aligned}{c}\int {{\rm{cs}}{{\rm{c}}^{\rm{4}}}} {\rm{(4x)dx = }}\int {{\rm{cs}}{{\rm{c}}^{\rm{2}}}} {\rm{(4x)cs}}{{\rm{c}}^{\rm{2}}}{\rm{(4x)dx}}\\{\rm{ = }}\int {{\rm{cs}}{{\rm{c}}^{\rm{2}}}} {\rm{(4x)}}\left( {{\rm{1 + co}}{{\rm{t}}^{\rm{2}}}{\rm{(4x)}}} \right){\rm{dx}}\\{\rm{ = }}\int {{\rm{cs}}{{\rm{c}}^{\rm{2}}}} {\rm{(4x)dx + }}\int {{\rm{cs}}{{\rm{c}}^{\rm{2}}}} {\rm{(4x)co}}{{\rm{t}}^{\rm{2}}}{\rm{(4x)dx}}\end{aligned}\)

The first component is

\(\int {{\rm{cs}}{{\rm{c}}^{\rm{2}}}} {\rm{(4x)dx = - }}\frac{{\rm{1}}}{{\rm{4}}}{\rm{cot(4x) + C}}\)

When the following formula is used for the elementary integral:

\(\int {{\rm{cs}}{{\rm{c}}^{\rm{2}}}} {\rm{xdx = - cotx + C}}{\rm{.}}\)

Evaluate the equation.

Next, figure out the second integral. Substitute\({\rm{t = cot4x}}\). Thus,

\(\begin{aligned}{c}\int {{\rm{cs}}{{\rm{c}}^{\rm{2}}}} {\rm{(4x)co}}{{\rm{t}}^{\rm{2}}}{\rm{(4x)dx = }}\left( {\begin{aligned}{*{20}{c}}{{\rm{t = cot(4x)}}}\\{{\rm{dt = - 4cs}}{{\rm{c}}^{\rm{2}}}{\rm{(4x)dx}}}\end{aligned}} \right)\\{\rm{ = - }}\frac{{\rm{1}}}{{\rm{4}}}\int {{{\rm{t}}^{\rm{2}}}} {\rm{dt = - }}\frac{{\rm{1}}}{{\rm{4}}}{\rm{ \times }}\frac{{{{\rm{t}}^{\rm{3}}}}}{{\rm{3}}}{\rm{ + C}}\\{\rm{ = - }}\frac{{\rm{1}}}{{{\rm{12}}}}{{\rm{t}}^{\rm{3}}}{\rm{ + C}}\\{\rm{ = - }}\frac{{\rm{1}}}{{{\rm{12}}}}{\rm{co}}{{\rm{t}}^{\rm{3}}}{\rm{(4x) + C}}\end{aligned}\)

Therefore, the integral value of the given equation is\(\int {{\rm{cs}}{{\rm{c}}^{\rm{4}}}{\rm{4xdx}}} {\rm{ = - }}\frac{{\rm{1}}}{{\rm{4}}}{\rm{cot(4x) - }}\frac{{\rm{1}}}{{{\rm{12}}}}{\rm{co}}{{\rm{t}}^{\rm{3}}}{\rm{(4x) + C}}\).