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In preparing to solve the first of those integrals we may make the trigonometric substitution sinxdx = -cosxdx so that we can integrate by parts. We have, then,

(Eq'n 1)

which gives us

(Eq'n 2)

We then obtain the solution of the second integral of our title by exploiting the associative law:

(Eq'n 3)

so we have finally

(Eq'n 4)

Example Application:

Electrical Energy from an Alternating Current:

If we establish a voltage difference of V=V0Sinωt across a resistance R, then the current I = V/R = V0Sinωt flows through the resistor and dissipates power at the rate P = VI = V02Sin2ωt/R. Over the time interval t the resistor dissipates energy in the amount equal to

(Eq'n 5)

in which we take ωt in the sine and the cosine to represent a particular phase angle by which the final state of the voltage differs from the initial state.


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