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Given the function f(x)=integral(0down-x up) (sint+cost)sint/(cost)^2,what is f(pie/4)?

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greenbel | Honors

Posted August 19, 2013 at 1:15 PM via web

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Given the function f(x)=integral(0down-x up) (sint+cost)sint/(cost)^2,what is f(pie/4)?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted August 19, 2013 at 2:02 PM (Answer #1)

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You need to evaluate the definite integral of the given function, hence, you need first replace `pi/4` for `x` , such that:

`f(pi/4) = int_0^(pi/4) ((sin t + cos t)*sin t)/(cos^2 t) dt`

You need to perform the multiplication to numerator, such that:

`f(pi/4) = int_0^(pi/4)(sin^2 t + sin t*cos t)/(cos^2 t) dt`

You need to use the property of linearity of integral, such that:

`f(pi/4) = int_0^(pi/4) sin^2 t/(cos^2 t) dt + int_0^(pi/4) sin t*cos t/(cos^2 t) dt`

`f(pi/4) = int_0^(pi/4) tan^2 t dt + int_0^(pi/4) tan tdt`

You need to use the following trigonometric identity:

`tan^2 t + 1 = 1/(cos^2 t)`

`f(pi/4) = int_0^(pi/4) (tan^2 t + 1 - 1) dt + int_0^(pi/4) tan tdt`

`f(pi/4) = int_0^(pi/4) (tan^2 t + 1)dt -int_0^(pi/4) dt +int_0^(pi/4) tan tdt`

Replacing `1/(cos^2 t)` for `tan^2 t + 1` yields:

`f(pi/4) = int_0^(pi/4)(1/(cos^2 t)) dt - t|_0^(pi/4) + int_0^(pi/4) sin t/cos t dt`

Since `1/(cos^2 t) = (tan t)'` yields:

`f(pi/4) = int_0^(pi/4)(tan t)'dt - t|_0^(pi/4) + int_0^(pi/4) sin t/cos t dt`

`f(pi/4) = tan t|_0^(pi/4) - t|_0^(pi/4) + int_0^(pi/4) sin t/cos t dt`

You need to use substitution to evaluate the definite integral `int_0^(pi/4) sin t/cos t dt` , such that:

`cos t = u => -sin tdt = du => sin tdt = -du`

`t = 0 => cos 0 = 1`

`t = pi/4 => cos (pi/4) = sqrt2/2`

Changing the variable yields:

`int_1^(sqrt2/2) -(du)/u = int_(sqrt2/2)^1 (du)/u = ln u|_(sqrt2/2)^1`

Using the fundamental theorem of calculus, yields:

`f(pi/4) = tan (pi/4) - tan 0 - pi/4 + 0 + ln 1 - ln (sqrt2/2)`

`f(pi/4) = 1 - 0 - pi/4 + 0 + 0 - ln (sqrt2/2)`

`f(pi/4) =1 - pi/4 - ln (sqrt2/2)`

Hence, evaluating the given definite integral, under the given conditions, yields `f(pi/4) =1 - pi/4 - ln (sqrt2/2).`

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