# Video: Determining the Definite Integral of a Trigonometric Function

Determine β«_(βπ) ^(βπ/4) 8 cos 5π dπ.

04:50

### Video Transcript

Determine the integral of eight cos five π dπ between negative π by four and negative π.

Our first step here is to take out the constant eight. This leaves us with eight multiplied by the integral of cos five π dπ. The integral of cos π is a standard one that we should know. It is equal to sin π, as integrating is the opposite, or inverse, of differentiating.

We want to integrate cos five π. This means we need to use another general rule. This states that the integral of cos ππ is equal to one over π multiplied by sin π. We differentiate the ππ to give us π. The integral of cos five π is equal to one-fifth multiplied by sin five π. We need to multiply this by eight and have limits of negative π by four and negative π. As with the eight at the start, we can take the constant one-fifth outside of the bracket. This gives us eight-fifths multiplied by sin five π.

Our next step is to substitute in our two limits and subtract the answers. Substituting in the upper limit gives us sin of five multiplied by negative π over four. This can be rewritten as sin of negative five π over four. Substituting in our lower limit gives us sin of five multiplied by negative π. Once again, this can be rewritten as sin of negative five π.

At this point, it is worth drawing the sine curve to see if negative five π by four and negative five π correspond with any our known angles. The sine curve has a maximum value of one and a minimum value of negative one. It has key values on the π-, or π₯-axis, of π by two and π, negative π by two and negative π. If you prefer to think of these angles in degrees. Itβs worth remembering that π radians is equal to 180 degrees.

The sine curve looks as shown in the diagram. However, at the moment, we have a slight problem as our two angles negative five π by four and negative five π donβt fit in the range. As the sine curve has a period of two π, it repeats every two π radians, we can continue the graph as shown.

We can see clearly from the graph that the sin of negative five π is equal to zero. Negative five π by four is shown in the diagram. By going vertically upwards to the sine curve and then horizontally along to the π¦-axis, we can see what this value will take. Due to the symmetry of the sine curse, sin of negative five π by four is equal to sin of π by four.

π by four is equal to 45 degrees and is one of our known angles. This is equal to root two over two. The sin of 45 degrees equals root two over two. Therefore, the sin of π by four radians must also equal root two over two. The sin of negative five π by four is equal to root two over two. And the sin of negative five π is equal to zero.

Root two over two minus zero is root two over two. So, we need to multiply this by eight-fifths. Multiplying the numerators gives us eight root two. And multiplying the denominators, gives us 10. We have eight root two over 10. Eight and 10 have a common factor of two. So, we can divide the numerator and denominator by two. This gives us four root two over five. We can, therefore, say that the integral of eight cos five π dπ between negative π by four and negative π is equal to four root two over five.