### Video Transcript

Calculate the integral the integral
of the sin of five π‘ π’ plus the cos of three π‘ π£ with respect to π‘.

The question wants us to calculate
the indefinite integral of a vector-valued function. We input a value of π‘ and it
outputs a position vector. We do this by integrating each of
our component functions with respect to π‘. Thatβs the integral of the sin of
five π‘ and the integral of the cos of three π‘. So we want to calculate the
integral of the sin of five π‘ with respect to π‘ and the integral of the cos of
three π‘ with respect to π‘.

We recall for a constant π, theWE
integral of the sin of ππ‘ with respect to π‘ is equal to negative the cos of ππ‘
divided by π plus the constant of integration π. Using this, we can evaluate the
integral of the sin of five π‘ with respect to π‘. Itβs equal to negative the cos of
five π‘ divided by five plus the constant of integration we will call π one. We also know for any constant π,
the integral of the cos of ππ‘ with respect to π‘ is equal to the sin of ππ‘
divided by π plus the constant of integration π. Using this, we have the integral of
the cos of three π‘ with respect to π‘ is equal to the sin of three π‘ over three
plus the constant of integration we will call π two.

Since weβve now found the integral
of each of our component functions, we can write the integral of the sin of five π‘
π’ plus the cos of three π‘ π£ with respect to π‘ as negative the cos of five π‘
over five plus π one π’ plus sin of three π‘ over three plus π two π£. We could leave our answer like
this. However, removing the parentheses
and rearranging, we can see that our answer is equal to negative the cos of five π‘
over five π’ plus the sin of three π‘ over three π£ plus π one π’ plus π two
π£. However, π one and π two are just
constants of integration. So we could combine this entire
expression into a vector we will call π.

Therefore, weβve shown the integral
of our vector-valued function with respect to π‘ is equal to negative one-fifth the
cos of five π‘ π’ plus the sin of three π‘ over three π£ plus π.