Video Transcript
In this video, we will learn how to
find indefinite integrals of functions that result in reciprocal trigonometric
functions.
We recall first that the derivative
with respect to 𝑥 of the sec of 𝑥, or sec 𝑥, is equal to the sec of 𝑥 times the
tan of 𝑥, or sec 𝑥 tan 𝑥. Recalling that integration is the
inverse of differentiation, integrating both sides of this equation with respect to
𝑥 undoes the differentiation operation while also adding an arbitrary constant of
integration. We can quote this as a standard
result. The indefinite integral of sec 𝑥
tan 𝑥 with respect to 𝑥 is equal to sec 𝑥 plus an arbitrary constant 𝐶. In our first example, we will use
this formula to solve an indefinite integral.
Determine the indefinite integral
of nine tan three 𝑥 sec three 𝑥 with respect to 𝑥.
The integrand here contains the
product of a tangent and a secant, both of which have the same argument three
𝑥. We can recall the following
standard integral. The indefinite integral of sec 𝑥
tan 𝑥 with respect to 𝑥 is equal to sec of 𝑥 plus 𝐶. To use this result in this example,
we need to modify the argument three 𝑥 of the trigonometric functions. We can do this by making a
substitution. We’ll let 𝑢 equal three 𝑥.
It follows then that d𝑢 by d𝑥 is
equal to three. And whilst d𝑢 by d𝑥 is not a
fraction, we can treat it a little like one. So equivalently d𝑢 is equal to
three d𝑥, or one-third d𝑢 is equal to d𝑥. We can now perform this
substitution replacing three 𝑥 with 𝑢 and d𝑥 with one-third d𝑢 to obtain the
indefinite integral of nine tan 𝑢 sec 𝑢 one-third d𝑢. The constant in the integrand
simplifies to three, and we can then bring this factor of three out the front to
give three multiplied by the indefinite integral of tan 𝑢 sec 𝑢 with respect to
𝑢.
By the standard result, this
integrates to three sec 𝑢 plus a constant of integration 𝐶. All that remains is to undo our
substitution, so we need to replace 𝑢 with three 𝑥, which gives our final answer
to the problem. We’ve found that the indefinite
integral of nine tan three 𝑥 sec three 𝑥 with respect to 𝑥 is equal to three sec
three 𝑥 plus a constant of integration 𝐶.
In our next example, we will solve
an indefinite integral which requires us to simplify the integrand first.
Determine the indefinite integral
of negative eight sec seven 𝑥 multiplied by negative four cos squared seven 𝑥 plus
six tan seven 𝑥 with respect to 𝑥.
Let’s begin by distributing the
parentheses in the integrand. The integrand becomes 32 sec seven
𝑥 cos squared seven 𝑥 minus 48 sec seven 𝑥 tan seven 𝑥. Now since sec of seven 𝑥 is equal
to one over cos of seven 𝑥, we can replace the first term with 32 cos squared seven
𝑥 over cos of seven 𝑥. We can cancel a factor of cos seven
𝑥 from the numerator and denominator, so the first term simplifies to 32 cos seven
𝑥. And now we need to determine the
indefinite integral of 32 cos seven 𝑥 minus 48 sec seven 𝑥 tan seven 𝑥 with
respect to 𝑥.
The first term in the integrand
involves a cosine function, and the second term involves a product of a secant and
tangent function each with the same argument of seven 𝑥. We will therefore need to recall
the following standard integrals. The integral of cos 𝑥 with respect
to 𝑥 is equal to sin 𝑥 plus 𝐶. And the integral of sec 𝑥 tan 𝑥
with respect to 𝑥 is equal to sec 𝑥 plus 𝐶.
However, before we can apply these
formulas to solve the given indefinite integral, we need to modify the argument of
the trigonometric functions. To do this, we can perform a
substitution. We let 𝑢 equal seven 𝑥, which in
turn implies that d𝑢 by d𝑥 is equal to seven. Now d𝑢 by d𝑥 is not a fraction,
but we can treat it a little like one. So equivalently one-seventh d𝑢 is
equal to d𝑥. Performing the substitution, we
obtain the indefinite integral of 32 cos 𝑢 minus 48 sec 𝑢 tan 𝑢 one-seventh
d𝑢.
We can simplify this by splitting
the integrand and taking each constant factor out the front to give 32 over seven
multiplied by the indefinite integral of cos 𝑢 d𝑢 minus 48 over seven multiplied
by the indefinite integral of sec 𝑢 tan 𝑢 d𝑢. Applying the standard results, we
obtain 32 over seven sin 𝑢 plus a constant of integration 𝐶 one minus 48 over
seven sec 𝑢 plus a constant of integration 𝐶 two. We can combine the two constants of
integration into a single arbitrary constant 𝐶. Finally, we need to reverse the
substitution by replacing 𝑢 with seven 𝑥. And so we obtain our final answer,
which is 32 over seven sin seven 𝑥 minus 48 over seven sec seven 𝑥 plus 𝐶.
Let’s consider another example
involving the product of sec 𝑥 and tan 𝑥.
Determine the indefinite integral
of seven sec 𝑥 multiplied by tan 𝑥 minus five sec 𝑥 with respect to 𝑥.
Since we have a factored expression
in the integrand, we’ll begin by distributing the parentheses, which gives the
indefinite integral of seven sec 𝑥 tan 𝑥 minus 35 sec squared 𝑥 with respect to
𝑥. The first term is the product of a
secant and tangent function, and the second is the square of the secant
function. To solve this problem, we will need
to recall two standard integrals. The indefinite integral of sec 𝑥
tan 𝑥 with respect to 𝑥 is equal to sec 𝑥 plus 𝐶. And the indefinite integral of sec
squared 𝑥 with respect to 𝑥 is equal to tan 𝑥 plus 𝐶.
Separating the integrand and taking
the constant factors out the front of each integral, we obtain seven multiplied by
the indefinite integral of sec 𝑥 tan 𝑥 with respect to 𝑥 minus 35 multiplied by
the indefinite integral of sec squared 𝑥 with respect to 𝑥. We can now apply the standard
results to this problem. Integrating, we obtain seven sec 𝑥
plus a constant of integration 𝐶 one minus 35 tan 𝑥 plus a constant of integration
𝐶 two. Combining the two arbitrary
constants into a single constant 𝐶, we obtain our final answer. The indefinite integral of seven
sec 𝑥 multiplied by tan 𝑥 minus five sec 𝑥 with respect to 𝑥 is equal to seven
sec 𝑥 minus 35 tan 𝑥 plus 𝐶.
In the examples we’ve seen so far,
we’ve used the formula for indefinite integrals that give results involving the
secant and tangent functions. Let’s now consider another type of
indefinite integral involving a different reciprocal trigonometric function.
Recall that the derivative of the
csc of 𝑥, or csc 𝑥, with respect to 𝑥 is equal to negative the csc of 𝑥 times
the cot of 𝑥. Integrating both sides with respect
to 𝑥 as before, we obtain another standard integral. The indefinite integral of csc 𝑥
cot 𝑥 with respect to 𝑥 is equal to negative csc 𝑥 plus 𝐶. Notice the similarity between this
standard integral and the previous one. By replacing each trigonometric
function in the integral by its complement, we obtain the same result, with the
exception that we switch the negative sign to a positive one. This resemblance is not a
coincidence.
Recall that the complementary
trigonometric functions cosine, cotangent, and cosecant take the complementary
angles of their counterparts. In other words, cos of 𝑥 is equal
to sin of 𝜋 by two minus 𝑥, cot of 𝑥 is equal to tan of 𝜋 by two minus 𝑥, and
csc of 𝑥 is equal to sec of 𝜋 by two minus 𝑥. Returning to our standard integral
then, we can rewrite the indefinite integral of csc 𝑥 cot 𝑥 with respect to 𝑥 as
the indefinite integral of sec of 𝜋 by two minus 𝑥 tan of 𝜋 by two minus 𝑥 with
respect to 𝑥. We can then use the substitution 𝑢
equals 𝜋 by two minus 𝑥 such that d𝑢 by d𝑥 is equal to negative one or
equivalently negative d𝑢 is equal to d𝑥.
Making this substitution in the
integral above, we obtain that the integral of csc 𝑥 cot 𝑥 with respect to 𝑥 is
equal to the negative of the integral of sec 𝑢 tan 𝑢 with respect to 𝑢. Using the standard integral of sec
𝑢 tan 𝑢 with respect to 𝑢, we obtain negative sec 𝑢 plus 𝐶. And replacing 𝑢 with 𝜋 by two
minus 𝑥, we obtain negative sec of 𝜋 by two minus 𝑥 plus 𝐶. And finally, switching both the
complement argument and the complement function, we get negative csc 𝑥 plus 𝐶. The inclusion of the negative sign
compared to the two previous standard integrals results from the substitution of 𝑢
equals 𝜋 by two minus 𝑥, leading to negative d𝑢 equals d𝑥, which then carries
through the rest of the integration.
Let’s now consider an example of an
indefinite integral using this formula.
Determine the indefinite integral
of two csc three 𝑥 cot three 𝑥 with respect to 𝑥.
The given integrand is the product
of a cosecant and cotangent function, both with the argument of three 𝑥. We recall the standard integral of
the product of the cosecant and cotangent functions. The indefinite integral of csc of
𝑥 multiplied by cot of 𝑥 with respect to 𝑥 is equal to the negative csc of 𝑥
plus 𝐶. In the given integrand, the
argument for both functions is three 𝑥 rather than 𝑥. So in order to apply the standard
integral, we need to use a substitution.
We let 𝑢 equal three 𝑥, which in
turn implies that d𝑢 by d𝑥 is equal to three. And so one-third d𝑢 is equivalent
to d𝑥. Making this change of variable in
the integral, we get the indefinite integral of two csc 𝑢 cot 𝑢 one-third d𝑢. Taking the constant factor of
two-thirds out the front of the integral and then applying our standard result gives
negative two-thirds csc 𝑢 plus a constant of integration 𝐶. All that remains is to reverse the
substitution by replacing 𝑢 with three 𝑥.
And so we obtain our final
answer. The indefinite interval of two csc
three 𝑥 cot three 𝑥 with respect to 𝑥 is equal to negative two-thirds csc of
three 𝑥 plus 𝐶.
For our final formula, we want to
determine the indefinite integral of csc squared of 𝑥 with respect to 𝑥. To start, we recall that the
integral of sec squared 𝑥 with respect to 𝑥 is equal to tan 𝑥 plus 𝐶. As discussed earlier, we can obtain
the complementary counterpart to this result by replacing sec squared 𝑥 and tan 𝑥
with their complement, csc squared of 𝑥 and cot 𝑥. But then we must also include a
factor of negative one on the right-hand side.
This gives our final standard
integral. The indefinite integral of csc
squared of 𝑥 with respect to 𝑥 is equal to negative cot of 𝑥 plus 𝐶. If ever we need to integrate tan
squared 𝑥 or cot squared 𝑥, we can use the trigonometric identities tan squared of
𝑥 is equal to sec squared of 𝑥 minus one and cot squared of 𝑥 is equal to csc
squared 𝑥 minus one to rewrite both of these integrals in terms of the standard
integrals we already know.
If you can’t remember either of
these identities, they can each be derived from the Pythagorean identity of sin
squared 𝑥 plus cos squared 𝑥 is identically equal to one. In our final example, we will
evaluate an indefinite integral involving cot squared 𝑥 by first applying this
trigonometric identity and second by applying the formula for the antiderivative of
csc squared 𝑥.
Determine the indefinite integral
of negative five multiplied by the cot squared of four 𝑥 plus seven plus one with
respect to 𝑥.
To begin, since negative five is a
constant, we can take it outside the integrand. Next, since the argument of the
trigonometric function is four 𝑥 plus seven instead of 𝑥, we can make a
substitution. We can let 𝑢 equal four 𝑥 plus
seven. And it then follows that d𝑢 by d𝑥
is equal to four or equivalently one-quarter d𝑢 is equal to d𝑥. Making this change of variable in
the integral, we obtain negative five multiplied by the indefinite integral of cot
squared 𝑢 plus one one-quarter d𝑢, which we can write as negative five over four
multiplied by the indefinite integral of cot squared 𝑢 plus one with respect to
𝑢.
The integrand contains the square
of the cotangent function. The antiderivative of cot squared
𝑥 is not readily available, but we do know that cot squared 𝑥 can be expressed in
terms of csc squared 𝑥, which we do know the antiderivative of. We recall that cot squared of 𝑥 is
equal to csc squared of 𝑥 minus one. And this can be obtained from the
Pythagorean identity. Hence, the integrand becomes csc
squared of 𝑢 minus one plus one or simply csc squared of 𝑢. We can now proceed by recalling the
standard result that the indefinite integral of csc squared of 𝑥 with respect to 𝑥
is equal to negative cot of 𝑥 plus 𝐶.
Applying this result, we obtain
negative five over four multiplied by negative cot 𝑢 plus 𝐶. Reversing our substitution, so
replacing 𝑢 with four 𝑥 plus seven, we obtain our final answer, which is that the
indefinite integral of negative five multiplied by cot squared of four 𝑥 plus seven
plus one with respect to 𝑥 is equal to five over four multiplied by cot of four 𝑥
plus seven plus 𝐶.
Let’s now recap some of the key
points from this video. Firstly, the indefinite integral of
the product of the sec of 𝑥 and the tan of 𝑥 with respect to 𝑥 is equal to the
sec of 𝑥 plus 𝐶. To find the integral of a
complementary trigonometric function, we replace every function with its complement
function and change the sign of the antiderivative. For example, for the integral of
the complementary function of the function in the first key point, the indefinite
integral of the csc of 𝑥 multiplied by the cot of 𝑥 with respect to 𝑥 is equal to
the negative csc of 𝑥 plus 𝐶. And likewise, for the integral of
the csc squared of 𝑥 with respect to 𝑥, we use our knowledge of the standard
integral of the sec squared of 𝑥 with respect to 𝑥 to give the negative of the cot
of 𝑥 plus 𝐶.
And finally, standard results for
the integrals of the tan squared of 𝑥 and the cot squared of 𝑥 are not readily
available, but we can use the trigonometric identities tan squared 𝑥 plus one is
identically equal to sec squared 𝑥 and cot squared 𝑥 plus one is identically equal
to csc squared 𝑥 to express any integrals of tan squared 𝑥 or cot squared 𝑥 in
terms of the standard integrals of sec squared 𝑥 and csc squared 𝑥 that we already
know.
