Q. 44 Evaluate the integral:鈭�02蟺sin... [FREE SOLUTION]

Chapter 11: Q. 44 (page 872)

Evaluate the integral:
${鈭�}_{0}^{2 蟺} ||<\sin t, \cos t, t>|| d t$

Short Answer

Expert verified

${鈭�}_{0}^{2 蟺} ||<\sin t, \cos t, t>|| d t = 蟺 \sqrt{1 + 4 蟺^{2}} + \frac{1}{2} \sin h^{- 1} (2 蟺) - \frac{1}{2} \sin h^{- 1} (0)$

Step by step solution

Given Information

Consider ${鈭�}_{0}^{2 蟺} ||<\sin t, \cos t, t>|| d t$

The objective is to evaluate the definite integral from $0$ to $2 蟺$ .

Calculation

The integral of a Vector Function

$鈭� r (t) d t = 鈭� x (t) i d t + y (t) j d t + z (t) k d t = i 鈭� x (t) d t + j 鈭� y (t) d t + k 鈭� z (t) d t + c$

Where $c$ is a vector constant and has the form $c = (c_{1}, c_{2}, c_{3})$ for scalar constants $c_{1}, c_{2}$ and $c_{3}$ .

Now,

$鈭� ||\sin t, \cos t, t|| d t = 鈭� \sqrt{\sin^{2} t + \cos^{2} t + t^{2}} d t = 鈭� \sqrt{1 + t^{2}} d t = \frac{t}{2} \sqrt{1 + t^{2}} + \frac{1}{2} \sin h^{- 1} (t)$

Expression

Now,

${鈭�}_{0}^{2 蟺} ||<\sin t, \cos t, t>|| d t = \frac{t}{2} \sqrt{1 + t^{2}} + \frac{1}{2} \sin h^{- 1} {(t)}_{0}^{2 蟺} = \frac{2 蟺}{2} \sqrt{1 + {(2 蟺)}^{2}} + \frac{1}{2} \sin h^{- 1} (2 蟺) - \frac{1}{2} \sin h^{- 1} (0) = 蟺 \sqrt{1 + 4 蟺^{2}} + \frac{1}{2} \sin h^{- 1} (2 蟺 0 - \frac{1}{2} \sin h^{- 1} (0)$

Thus ${鈭�}_{0}^{2 蟺} ||<\sin t, \cos t, t>|| d t = 蟺 \sqrt{1 + 4 蟺^{2}} + \frac{1}{2} \sinh^{- 1} (2 蟺) - \frac{1}{2} \sin h^{- 1} (0)$

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Evaluate the integral:
${鈭�}_{0}^{2 蟺} ||<\sin t, \cos t, t>|| d t$

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Evaluate the integral:
${鈭�}_{0}^{2 蟺} ||<\sin t, \cos t, t>|| d t$