How do you evaluate the definite integral $\int (sin t + cos t) d t$ $int (sint+cost)dt$ from $[0, \frac{π}{4}]$ $[0,pi/4]$ ?

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How do you evaluate the definite integral $\int (sin t + cos t) d t$ $int (sint+cost)dt$ from $[0, \frac{π}{4}]$ $[0,pi/4]$ ?

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Answer 1 · 2016-11-06T02:29:26

$\int_{0}^{\frac{π}{4}} (sin t + cos t) d t = 1$ $int_0^(pi/4)(sint+cost)dt=1$

Explanation:

We have to know that:

$\int (f (x) + g (x)) d x = \int f (x) d x + \int g (x) d x$ $int(f(x)+g(x))dx=intf(x)dx+intg(x)dx$
$\int sin t d t = - cos t + C$ $intsintdt=-cost+C$
$\int cos t d t = sin t + C$ $intcostdt=sint+C$

Combining these and keeping the bounds of the integral, we see that:

$\int_{0}^{\frac{π}{4}} (sin t + cos t) d t = {[- cos t + sin t]}_{0}^{\frac{π}{4}}$ $int_0^(pi/4)(sint+cost)dt=[-cost+sint]_0^(pi/4)$

Evaluating at $t = \frac{π}{4}$ $t=pi/4$ then subtracting the function evaluated at $t = 0$ $t=0$ :

$= [- cos (\frac{π}{4}) + sin (\frac{π}{4})] - [- cos 0 + sin 0]$ $=[-cos(pi/4)+sin(pi/4)]-[-cos0+sin0]$

$= (- \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}) - (- 1 + 0)$ $=(-sqrt2/2+sqrt2/2)-(-1+0)$

$= 0 + 1$ $=0+1$

$= 1$ $=1$

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How do you evaluate the definite integral $\int (sin t + cos t) d t$ $int (sint+cost)dt$ from $[0, \frac{π}{4}]$ $[0,pi/4]$ ?

1 Answer

Explanation:

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