Video Transcript
Determine the integral of the sec
of two π₯ plus one over three multiplied by the tan of two π₯ plus one over three
with respect to π₯.
In this question, weβre asked to
evaluate the integral of a product of two trigonometric functions, the secant of an
argument and the tangent of an argument. And this can immediately remind us
of one of our standard integral results. The integral of the sec of π
multiplied by the tan of π with respect to π is the sec of π plus the constant of
integration πΆ. We canβt apply this result directly
to this integral because we can see the argument of the secant function and the
tangent function is more complicated. So instead, weβll integrate this by
using a substitution. Weβll set π’ equal to the argument
two π₯ plus one over three.
To apply integration by
substitution, we recall we need to find an expression for the differentials. So weβre going to need to
differentiate our substitution π’ with respect to π₯. And since π’ is a linear function,
the derivative of π’ with respect to π₯ will be the coefficient of π₯. In this case, dπ’ by dπ₯ is equal
to two-thirds. Now dπ’ by dπ₯ is not a
fraction. However, we can treat it a little
bit like a fraction to help us find an expression for the differentials. We get that dπ₯ is equal to three
over two dπ’.
Weβre now ready to apply our
substitution to help us evaluate our integral. Weβll replace two π₯ plus one over
three with π’ and dπ₯ with three over two dπ’. This gives us the integral of the
sec of π’ times the tan of π’ times three over two with respect to π’. And now this is almost in the form
of our integral result. We just need to take the factor of
three over two outside of our integral. We get three over two times the
integral of sec π’ times tan π’ with respect to π’.
We then apply our integral
rule. We get three over two times the sec
of π’ plus the constant of integration πΆ, where itβs important to remember that πΆ
is just a constant. So itβs not affected by the factor
of three over two. Finally, since our original
integral was in terms of π₯, we should rewrite this expression in terms of π₯. We need to use our π’-substitution,
which gives us three over two times the sec of two π₯ plus one over three plus π,
which is our final answer.
Therefore, by using a
π’-substitution and one of our standard integral results, we were able to show the
integral of the sec of two π₯ plus one over three times the tan of two π₯ plus one
over three with respect to π₯ is equal to three over two times the sec of two π₯
plus one over three plus the constant of integration πΆ.
