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The value of $\int^{\pi/4} _0 x \cos(x^2) dx$ correct to three decimal places (assuming that $\pi = 3.14$) is ____
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Best answer
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$\int _{0}^{\frac{\pi}{4} } x \cos (x^{2}) dx$

put x$^{2} = t$

$2x dx = dt $

$t$ will range from  $0$ to $\frac{\pi ^{2} }{16}$

Now our new integral is : $\frac{1}{2}\int _{0}^{\frac{\pi ^{2} }{16}} \cos(t) dt$

= $\frac{1}{2} [\sin (t) ]_{0}^{\frac{\pi ^{2} }{16}} = \frac{1}{2} [\sin (.616225)- 0]$ = $\frac{0.5779}{2}$

 

 0.289 Answer 

edited by
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7 votes

Let ${x^2}$=t, So, 2xdx=dt

when ${x^2}$=0, t =0 and when ${x^2}$=${\Pi /4}$, t= $\Pi ^{2}/16$

Substituting it in the question, the integral becomes 

$\int_{0}^{\Pi /4}$cos(t)/2 dt =$[sin t/2]_{0}^{\Pi^{2}/16}$

=sin(0.617 radians)-0 / 2

=0.5785909/2

=0.2892 (ANS)

0 votes
0 votes

0.2892

Solved by letting x=t

Answer:

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