# What is `int_0^pi f(x)?` f(x)=(sin x)^3cosx

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the definite integral, such that:

`int_0^pi f(x)dx = int_0^pi sin^3 x*cos x dx = int_0^pi sin^2 x*cos x*sin x dx`

You need to use the basic trigonometric identity, such that:

`sin^2 x = 1 - cos^2 x`

`int_0^pi (1 - cos^2 x)*cos x*sin x dx = int_0^pi cos x*sin x dx - int_0^pi cos^3 x*sin x dx`

You should come up with the following substitution, such that:

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

Changing the limits of integration, yields:

`x = 0 => cos 0 = 1 = t`

`x = pi => cos pi = -1 = t`

Replacing the variable, yields:

`int_1^(-1) t*(-dt) - int_1^(-1) t^3*(-dt)`

Using the following property of definite integrals yields:

`int_a^b f(x)dx = -int_b^a f(x)dx`

Reasoning by analogy, yields:

`int_(-1)^1 t dt - int_(-1)^1 t^3 dt = t^2/2|_(-1)^1 - t^4/4|_(-1)^1`

Using the fundamental theorem of calculus, yields:

`int_(-1)^1 t dt - int_(-1)^1 t^3 dt = 1/2 - (-1)^2/2 - 1/4 + (-1)^4/4`

`int_(-1)^1 t dt - int_(-1)^1 t^3 dt = 1/2 - 1/2 - 1/4 + 1/4`

`int_(-1)^1 t dt - int_(-1)^1 t^3 dt = 0`

Hence, evaluating the given definite intgeral, using the indicated substitutions, yields `int_0^pi sin^3 x*cos x dx = 0` .