Video Transcript
Evaluate the integral the integral from zero to 𝜋 over six of three times the cos of three 𝑡 𝐢 plus four times the sec squared of two 𝑡 𝐣 with respect to 𝑡.
The question wants us to calculate the definite integral of a vector-valued function. To do this, we just calculate the definite integral of each of the component functions. We’ll start with the definite integral of our first component function. That’s the integral from zero to 𝜋 by six of three times the cos of three 𝑡 with respect to 𝑡.
To integrate this, we recall, for any constants 𝑎 and 𝑛, where 𝑛 is not equal to zero, the integral of 𝑎 cos of 𝑛𝑡 with respect to 𝑡 is equal to eight times the sin of 𝑛𝑡 divided by 𝑛 plus the constant of integration 𝑐. Using this to evaluate our integral, we get three times the sin of three 𝑡 all divided by three evaluated at the limits of our integral 𝑡 is equal to zero and 𝑡 is equal to 𝜋 by six.
We don’t include the constant of integration because we’re calculating a definite integral. We can simplify this. The shared factor of three in our numerator and our denominator cancel. Evaluating this at the limits of our integral gives us the sin of three times 𝜋 by six minus the sin of three times zero.
We have three times 𝜋 by six is equal to 𝜋 by two and three times zero is equal to zero. We know that sin of 𝜋 by two is equal to one and the sin of zero is equal to zero. So this evaluates to just give us one.
We now need to find the definite integral of our second component function. That’s the integral from zero to 𝜋 by six of four times the sec squared of two 𝑡 with respect to 𝑡. To integrate this, we recall one of our standard integral rules for trigonometric functions. For constants 𝑎 and 𝑛, where 𝑛 is not equal to zero, the integral of eight times the sec squared of 𝑛𝑡 with respect to 𝑡 is equal to 𝑎 times the tan of 𝑛𝑡 over 𝑛 plus the constant of integration 𝑐.
Using this to evaluate our integral, we get four times the tan of two 𝑡 divided by two evaluated at the limits of our integral 𝑡 is equal to zero and 𝑡 is equal to 𝜋 by six. Again, we don’t include the constant of integration because we’re calculating a definite integral.
We can also simplify this. Four divided by two is just equal to two. Evaluating this at the limits of our integral gives us two times the tan of two times 𝜋 by six minus two times the tan of two times zero. This simplifies to give us two times the tan of 𝜋 by three minus two times the tan of zero. The tan of 𝜋 by three is equal to the square root of three. This is a standard trigonometric result we should commit to memory. We multiply this by two, and then we subtract two times the tan of zero, which is just equal to zero. So the definite integral of our second component function gave us two times the square root of three.
Now that we’ve evaluated the definite integral of both of our component functions. We can conclude the integral from zero to 𝜋 by six of three times the cos of three 𝑡 𝐢 plus four times the sec squared of two 𝑡 𝐣 with respect to 𝑡 is equal to one 𝐢 plus two root three 𝐣. And we can simplify one 𝐢 to just be 𝐢.
