1, = ſtan
x dx , then
1
In
tan
n-1 x-In-2
n-1
Answers
Answer:
Step-by-step explanation:
Functions consisting of products of the sine and cosine can be integrated by using substitution and trigonometric identities. These can sometimes be tedious, but the technique is
straightforward. Some examples will suffice to explain the approach.
EXAMPLE 10.1.1 Evaluate Z
sin5 x dx. Rewrite the function:
Z
sin5 x dx =
Z
sin x sin4 x dx =
Z
sin x(sin2 x)
2
dx =
Z
sin x(1 − cos2 x)
2
dx.
Now use u = cos x, du = − sin x dx:
Z
sin x(1 − cos2 x)
2
dx =
Z
−(1 − u
2
)
2
du
=
Z
−(1 − 2u
2 + u
4
) du
= −u +
2
3
u
3 −
1
5
u
5 + C
= − cos x +
2
3
cos3 x −
1
5
cos5 x + C.
function:
Z
sin6 x dx =
Z
(sin2 x)
3
dx =
Z
(1 − cos 2x)
3
8
dx
=
1
8
Z
1 − 3 cos 2x + 3 cos2
2x − cos3
2x dx.
Now we have four integrals to evaluate:
Z
1 dx = x
and
Z
−3 cos 2x dx = −
3
2
sin 2x
are easy. The cos3
2x integral is like the previous example:
Z
− cos3
2x dx =
Z
− cos 2x cos2
2x dx
=
Z
− cos 2x(1 − sin2
2x) dx
=
Z
−
1
2
(1 − u
2
) du
= −
1
2
u −
u
3
3
= −
1
2
sin 2x −
sin3
2x
3
.
And finally we use another trigonometric identity, cos2 x = (1 + cos(2x))/2:
Z
3 cos2
2x dx = 3 Z
1 + cos 4x
2
dx =
3
2
x +
sin 4x
4