Question

Evaluate the integral.

\(\int {\left( {t{e^{2t}}i + \frac{t}{{1 - t}}j + \frac{1}{{\sqrt {1 - {t^2}} }}k} \right)} dt\)

Short answer

Therefore, the solution is \(\int {\left( {t{{\rm{e}}^{2t}}{\bf{i}} + \frac{t}{{1 - t}}{\bf{j}} + \frac{1}{{\sqrt {1 - {t^2}} }}{\bf{k}}} \right)} dt = {\bf{i}}\frac{{{{\rm{e}}^{2t}}(2t - 1)}}{4} + {\bf{j}}( - t - \ln |t - 1|) + {\bf{k}}{\sin ^{ - 1}}t + \vec C\)

Step-by-step solution

Step 1: Given information

The given value is:

\(\int {\left( {t{e^{2t}}{\bf{i}} + \frac{t}{{1 - t}}{\bf{j}} + \frac{1}{{\sqrt {1 - {t^2}} }}{\bf{k}}} \right)} dt\)

Evaluate the provided integral

Integrate the given integral component-wise, that is,

\(\int {\left( {t{{\rm{e}}^{2t}}{\bf{i}} + \frac{t}{{1 - t}}{\bf{j}} + \frac{1}{{\sqrt {1 - {t^2}} }}{\bf{k}}} \right)} dt = {\bf{i}}\int t {{\rm{e}}^{2t}}dt + {\bf{j}}\int {\frac{t}{{1 - t}}} dt + {\bf{k}}\int {\frac{1}{{\sqrt {1 - {t^2}} }}} dt\)

Now, according to integration by parts rule, we know that:

\(\int f \cdot {g^\prime } = fg - \int {{f^\prime }} \cdot g\)

\(f(t) = t,{g^\prime }(t) = {{\rm{e}}^{2t}}\)

\({f^\prime }(t) = 1,g(t) = \frac{{{{\rm{e}}^{2t}}}}{2}\)

This implies,

\({\bf{i}}\int t {{\rm{e}}^{2t}}dt = t \cdot \frac{{{{\rm{e}}^{2t}}}}{2}{\bf{i}} - \int 1 \cdot \frac{{{{\rm{e}}^{2t}}}}{2} \cdot dt \cdot {\bf{i}} = t \cdot \frac{{{{\rm{e}}^{2t}}}}{2}{\bf{i}} - \frac{{{{\rm{e}}^{2t}}}}{4}{\bf{i}} = \frac{{{{\rm{e}}^{2t}}(2t - 1)}}{4}{\bf{i}}\)

Simplify the function

Now, subtract and add 1 to simplify the function and use table of the elementary integrals,

\({\bf{j}}\int {\frac{t}{{1 - t}}} dt = {\bf{j}}\int {\frac{{(t - 1) + 1}}{{1 - t}}} dt\)

\( = {\bf{j}}\int {\left( {\frac{{t - 1}}{{1 - t}} + \frac{1}{{1 - t}}} \right)} dt\)

\( = {\bf{j}}\int {\left( { - 1 + \frac{1}{{1 - t}}} \right)} dt\)

\( = {\bf{j}}\int {\left( { - 1 - \frac{1}{{t - 1}}} \right)} dt\)

\( = {\bf{j}}\left( { - t - \ln |t - 1| + {c_2}} \right)\)

The next component is,

\({\bf{k}}\int {\frac{1}{{\sqrt {1 - {t^2}} }}} dt = {\bf{k}}\left( {{{\sin }^{ - 1}}t + {c_3}} \right)\)

Combining all three components together, we get,

\(\vec C = \left( {{c_1},{c_2},{c_3}} \right) = i{c_1} + j{c_2} + k{c_3}\)

\(\int {\left( {t{{\rm{e}}^{2t}}{\bf{i}} + \frac{t}{{1 - t}}{\bf{j}} + \frac{1}{{\sqrt {1 - {t^2}} }}{\bf{k}}} \right)} dt = {\bf{i}}\frac{{{{\rm{e}}^{2t}}(2t - 1)}}{4} + {\bf{j}}( - t - \ln |t - 1|) + {\bf{k}}{\sin ^{ - 1}}t + \vec C\)