How calculate int_(pi/3)^(pi/2) 1/sin x dx ?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You may re-write the integrand, such that:

`1/sin x = (sin x)/(sin^2 x) `

Using Pythagorean identity, yields:

`(sin x)/(sin^2 x) = (sin x)/(1 - cos^2 x)`

You should use integration by substitution, hence, you should come up with the following substitution, such that:

`cos x = t => - sin x dx = dt => sin x dx = -dt`

Evaluating the limits of integration, yields:

`cos (pi/3) = 1/2 = t`

`cos (pi/2) = 0 = t`

Changing the variable yields:

`int_(pi/3)^(pi/2) 1/sin x dx = int_(1/2)^0 (-dt)/(1 - t^2)`

The sign - changes the order of limits of integration, such that:

`int_(1/2)^0 (-dt)/(1 - t^2) = int_0^(1/2) (dt)/(1 - t^2)`

`int_0^(1/2) (dt)/(1 - t^2) = -(1/2)ln|(t-1)/(t+1)||_0^(1/2)`

By fundamental theorem of calculus, yields:

`int_0^(1/2) (dt)/(1 - t^2) = -(1/2)(ln|(1/2-1)/(1/2+1)| - ln|-1|)`

`int_0^(1/2) (dt)/(1 - t^2) = -(1/2)(ln(1/3))`

Using the power property of logarithms yields:

`int_0^(1/2) (dt)/(1 - t^2) = -(ln(1/3)^(1/2))`

`int_0^(1/2) (dt)/(1 - t^2) = ln 1/(sqrt3)`

`int_0^(1/2) (dt)/(1 - t^2) = -ln (sqrt3/3)`

Hence, evaluating the definite integral, using integration by substitution, yields `int_(pi/3)^(pi/2) 1/sin x dx = -ln (sqrt3/3).`

Sources:
aruv's profile pic

aruv | High School Teacher | (Level 2) Valedictorian

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`1/(sin(x))=cosec(x)`

`d/(dx)(cosec(x)+cot(x))=-cosec(x)cot(x)-cosec^2(x)`

`=-cosec(x)(cosec(x)+cot(x))`

Thus

`int_(pi/3)^(pi/2)1/(sin(x))dx=int_(pi/3)^(pi/2)cosec(x)dx`

`=int_(pi/3)^(pi/2)(cosec(x)(cosec(x)+cot(x)))/(cosec(x)+cot(x))dx`

`=-ln(cosex(x)+cot(x))}_(pi/3)^(pi/2)`

`=-{ln(1+0)-ln(2/sqrt(3)+1/sqrt(3))}`

`=ln(3/sqrt(3))`

`=ln(sqrt(3))`

`=(1/2)ln(3)`

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