Question

Evaluate the integral.

\(\int_0^{\pi /2} {\left( {3si{n^2}tcosti + 3sintco{s^2}tj + 2sintcostk} \right)} dt\)

Short answer

Therefore, the solution is \(\int_0^{\frac{\pi }{2}} {\left( {3{{\sin }^2}t\cos t{\bf{i}} + 3\sin t{{\cos }^2}t{\bf{j}} + 2\sin t\cos t{\bf{k}}} \right)} dt = {\bf{i}} + {\bf{j}} + {\bf{k}}\)

Step-by-step solution

Step 1: Given information

The given value is:

\(\int_0^{\pi /2} {\left( {3{{\sin }^2}t\cos t{\bf{i}} + 3\sin t{{\cos }^2}t{\bf{j}} + 2\sin t\cos t{\bf{k}}} \right)} dt\)

Step 2: Evaluate the provided integral

Let us evaluate the given integral component-wise as shown below,\(\int_0^{\frac{\pi }{2}} {\left( {3si{n^2}tcosti + 3sintco{s^2}tj + 2sintcostk} \right)} dt - - - - (1)\)

Using the fact that the integral of a vector function\({\bf{r}}(t)\)is a vector obtained by integrating the components of\({\bf{r}}(t)\)we get:

\(\begin{array}{l}\int_0^{\frac{\pi }{2}} {\left( {3{{\sin }^2}t\cos t{\bf{i}} + 3\sin t{{\cos }^2}t{\bf{j}} + 2\sin t\cos t{\bf{k}}} \right)} dt = \left( {3\int_0^{\frac{\pi }{2}} {{{\sin }^2}} t\cos tdt} \right)i + \\{\rm{ }} + \left( {3\int_0^{\frac{\pi }{2}} {\sin } t{{\cos }^2}tdt} \right){\bf{j}} + \\{\rm{ }} + \left( {2\int_0^{\frac{\pi }{2}} {\sin } t\cos tdt} \right){\bf{k}}\end{array}\)

Let us denote the first integral by \({S_1},\) second by \({S_2}\) and the third by \({S_3}.\)

Step 3: The following is the result of using the parts-based integration method

\({S_1} = \int_0^{\frac{\pi }{2}} {{{\sin }^2}} t\cos tdt = \left| {\begin{array}{*{20}{c}}u& = &{{{\sin }^2}t}\\{du}& = &{2\sin t\cos tdt}\\v& = &{\int {\cos } tdt = \sin t}\end{array}} \right|\)

\({S_1} = \left. {\left( {{{\sin }^2}t \cdot \sin t} \right)} \right|_0^{\frac{\pi }{2}} - \int_0^{\frac{4}{2}} {\sin } t \cdot 2\sin t\cos tdt\)

\({S_1} = \left. {{{\sin }^3}t} \right|_0^{\frac{\pi }{2}} - 2\underbrace {\int_0^{\frac{\pi }{2}} {{{\sin }^2}} t\cos tdt}_{{S_1}}\)

\({S_1} = {\sin ^3}\frac{\pi }{2} - {\sin ^3}(0) - 2{S_1}\)

\(3{S_1} = 1 \Rightarrow {S_1} = \frac{1}{3}\)

\({S_2} = \int_0^{\frac{\pi }{2}} {\sin } t{\cos ^2}tdt = \left| {\begin{array}{*{20}{c}}u& = &{{{\cos }^2}t}\\{du}& = &{ - 2\cos t\sin tdt}\\v& = &{\int {\sin } tdt = - \cos t}\end{array}} \right|\)

\({S_2} = - \left. {{{\cos }^3}t} \right|_0^{\frac{\pi }{2}} + \int_0^{\bar 2} {\cos } t \cdot ( - 2\cos t\sin t)dt\)

\({S_2} = - \left. {{{\cos }^3}t} \right|_0^{\frac{\pi }{2}} - 2\underbrace {\int_0^{\frac{\pi }{2}} {{{\cos }^2}} t\sin tdt}_{{S_2}}\)

\({S_2} = - {\cos ^3}\frac{\pi }{2} + {\cos ^3}(0) - 2{S_2}\)

\(3{S_2} = 1 \Rightarrow {S_2} = \frac{1}{{\bf{3}}}\)

Use substitution method

\({S_3} = \int_0^{\frac{\pi }{2}} {\sin } t\cos tdt = \left| {\begin{array}{*{20}{c}}{\sin t}& = &\alpha \\{\cos tdt}& = &{d\alpha }\\0& \to &0\\{\frac{\pi }{2}}& \to &1\end{array}} \right|\)

\({S_3} = \int_0^1 \alpha d\alpha = \left. {\frac{{{\alpha ^2}}}{2}} \right|_0^1 = \frac{1}{2}\)

Substituting\({S_1},{S_2}\)and\({S_3}\)in (1) we get:

\(\int_0^{\frac{\pi }{2}} {\left( {3{{\sin }^2}t\cos t{\bf{i}} + 3\sin t{{\cos }^2}t{\bf{j}} + 2\sin t\cos t{\bf{k}}} \right)} dt = 3 \cdot \frac{1}{3}{\bf{i}} + 3 \cdot \frac{1}{3}{\bf{j}} + 2 \cdot \frac{1}{2}{\bf{k}}\)

\( = i + j + k\)

Notice that the integrals\({S_1}\)and\({S_2}\)can also be computed using the

substitution method.