Integrals with tan x and sec x
Z
tan3 xdx
Integrals with tan x and sec x
Z
We want to use the identity:
tan3 xdx
Integrals with tan x and sec x
Z
tan3 xdx
We want to use the identity:
tan2 x = sec2 x − 1
Integrals with tan x and sec x
Z
tan3 xdx
We want to use the identity:
tan2 x = sec2 x − 1
So we write the integral as:
Integrals with tan x and sec x
Z
tan3 xdx
We want to use the identity:
tan2 x = sec2 x − 1
So we write the integral as:
Z
tan3 xdx =
Integrals with tan x and sec x
Z
tan3 xdx
We want to use the identity:
tan2 x = sec2 x − 1
So we write the integral as:
Z
3
tan xdx =
Z
Integrals with tan x and sec x
Z
tan3 xdx
We want to use the identity:
tan2 x = sec2 x − 1
So we write the integral as:
Z
Z
3
tan xdx = tan2 x
Integrals with tan x and sec x
Z
tan3 xdx
We want to use the identity:
tan2 x = sec2 x − 1
So we write the integral as:
Z
Z
3
tan xdx = tan2 x tan x
Integrals with tan x and sec x
Z
tan3 xdx
We want to use the identity:
tan2 x = sec2 x − 1
So we write the integral as:
Z
Z
3
tan xdx = tan2 x tan xdx
Integrals with tan x and sec x
Z
tan3 xdx
We want to use the identity:
tan2 x = sec2 x − 1
So we write the integral as:
Z
Z
3
tan xdx = tan2 x tan xdx =
Integrals with tan x and sec x
Z
tan3 xdx
We want to use the identity:
tan2 x = sec2 x − 1
So we write the integral as:
Z
Z
Z
3
2
tan xdx = tan x tan xdx =
Integrals with tan x and sec x
Z
tan3 xdx
We want to use the identity:
tan2 x = sec2 x − 1
So we write the integral as:
Z
Z
Z
3
2
tan xdx = tan x tan xdx =
sec2 x − 1
Integrals with tan x and sec x
Z
tan3 xdx
We want to use the identity:
tan2 x = sec2 x − 1
So we write the integral as:
Z
Z
Z
3
2
tan xdx = tan x tan xdx =
sec2 x − 1 tan xdx
Integrals with tan x and sec x
Z
tan3 xdx
We want to use the identity:
tan2 x = sec2 x − 1
So we write the integral as:
Z
Z
Z
3
2
tan xdx = tan x tan xdx =
Z
=
sec2 x − 1 tan xdx
Integrals with tan x and sec x
Z
tan3 xdx
We want to use the identity:
tan2 x = sec2 x − 1
So we write the integral as:
Z
Z
Z
3
2
tan xdx = tan x tan xdx =
Z
=
sec2 x
sec2 x − 1 tan xdx
Integrals with tan x and sec x
Z
tan3 xdx
We want to use the identity:
tan2 x = sec2 x − 1
So we write the integral as:
Z
Z
Z
3
2
tan xdx = tan x tan xdx =
Z
=
sec2 x tan x
sec2 x − 1 tan xdx
Integrals with tan x and sec x
Z
tan3 xdx
We want to use the identity:
tan2 x = sec2 x − 1
So we write the integral as:
Z
Z
Z
3
2
tan xdx = tan x tan xdx =
Z
=
sec2 x tan xdx
sec2 x − 1 tan xdx
Integrals with tan x and sec x
Z
tan3 xdx
We want to use the identity:
tan2 x = sec2 x − 1
So we write the integral as:
Z
Z
Z
3
2
tan xdx = tan x tan xdx =
Z
=
sec2 x − 1 tan xdx
sec2 x tan xdx−
Integrals with tan x and sec x
Z
tan3 xdx
We want to use the identity:
tan2 x = sec2 x − 1
So we write the integral as:
Z
Z
Z
3
2
tan xdx = tan x tan xdx =
Z
=
2
sec x tan xdx −
sec2 x − 1 tan xdx
Z
Integrals with tan x and sec x
Z
tan3 xdx
We want to use the identity:
tan2 x = sec2 x − 1
So we write the integral as:
Z
Z
Z
3
2
tan xdx = tan x tan xdx =
Z
=
2
sec x tan xdx −
sec2 x − 1 tan xdx
Z
tan x
Integrals with tan x and sec x
Z
tan3 xdx
We want to use the identity:
tan2 x = sec2 x − 1
So we write the integral as:
Z
Z
Z
3
2
tan xdx = tan x tan xdx =
Z
=
2
sec x tan xdx −
sec2 x − 1 tan xdx
Z
tan xdx
Integrals with tan x and sec x
Z
tan3 xdx
We want to use the identity:
tan2 x = sec2 x − 1
So we write the integral as:
Z
Z
Z
3
2
tan xdx = tan x tan xdx =
Z
=
sec2 x − 1 tan xdx
Z
2
sec x tan xdx −
Z
=
tan xdx
Integrals with tan x and sec x
Z
tan3 xdx
We want to use the identity:
tan2 x = sec2 x − 1
So we write the integral as:
Z
Z
Z
3
2
tan xdx = tan x tan xdx =
Z
=
sec2 x − 1 tan xdx
Z
2
sec x tan xdx −
Z
=
sec2 x
tan xdx
Integrals with tan x and sec x
Z
tan3 xdx
We want to use the identity:
tan2 x = sec2 x − 1
So we write the integral as:
Z
Z
Z
3
2
tan xdx = tan x tan xdx =
Z
=
Z
2
sec x tan xdx −
Z
=
sec2 x − 1 tan xdx
sec2 x tan x
tan xdx
Integrals with tan x and sec x
Z
tan3 xdx
We want to use the identity:
tan2 x = sec2 x − 1
So we write the integral as:
Z
Z
Z
3
2
tan xdx = tan x tan xdx =
Z
=
2
Z
sec x tan xdx −
Z
=
sec2 x − 1 tan xdx
tan xdx
sec2 x tan xdx
Integrals with tan x and sec x
Z
tan3 xdx
We want to use the identity:
tan2 x = sec2 x − 1
So we write the integral as:
Z
Z
Z
3
2
tan xdx = tan x tan xdx =
Z
=
Z
2
sec x tan xdx −
Z
=
sec2 x − 1 tan xdx
2
tan xdx
sec x tan xdx −
Z
Integrals with tan x and sec x
Z
tan3 xdx
We want to use the identity:
tan2 x = sec2 x − 1
So we write the integral as:
Z
Z
Z
3
2
tan xdx = tan x tan xdx =
Z
Z
sec x tan xdx −
=
Z
=
2
2
sec x tan xdx −
sec2 x − 1 tan xdx
tan xdx
Z
sin x
dx
cos x
Integrals with tan x and sec x
Integrals with tan x and sec x
Z
2
sec x tan xdx −
Z
sin x
dx
cos x
Integrals with tan x and sec x
Z
2
sec x tan xdx −
We make the substitutions:
Z
sin x
dx
cos x
Integrals with tan x and sec x
Z
Z
2
sec x tan xdx −
We make the substitutions:
u=
sin x
dx
cos x
Integrals with tan x and sec x
Z
2
Z
sec x tan xdx −
We make the substitutions:
u = tan x
sin x
dx
cos x
Integrals with tan x and sec x
Z
2
sec x tan xdx −
Z
sin x
dx
cos x
We make the substitutions:
u = tan x,
du
=
dx
Integrals with tan x and sec x
Z
2
Z
sec x tan xdx −
sin x
dx
cos x
We make the substitutions:
u = tan x,
du
= sec2 x
dx
Integrals with tan x and sec x
Z
Z
2
sec x tan xdx −
sin x
dx
cos x
We make the substitutions:
u = tan x,
v=
du
= sec2 x
dx
Integrals with tan x and sec x
Z
Z
2
sec x tan xdx −
sin x
dx
cos x
We make the substitutions:
u = tan x,
du
= sec2 x
dx
v = cos x
Integrals with tan x and sec x
Z
2
Z
sec x tan xdx −
sin x
dx
cos x
We make the substitutions:
u = tan x,
v = cos x,
du
= sec2 x
dx
dv
=
dx
Integrals with tan x and sec x
Z
2
Z
sec x tan xdx −
sin x
dx
cos x
We make the substitutions:
u = tan x,
du
= sec2 x
dx
v = cos x,
dv
= − sin x
dx
Integrals with tan x and sec x
Z
2
Z
sec x tan xdx −
sin x
dx
cos x
We make the substitutions:
u = tan x,
du
= sec2 x
dx
v = cos x,
dv
= − sin x
dx
Replacing in the integrals:
Integrals with tan x and sec x
Z
Z
2
sec x tan xdx −
sin x
dx
cos x
We make the substitutions:
u = tan x,
du
= sec2 x
dx
v = cos x,
dv
= − sin x
dx
Replacing in the integrals:
Z
u
|{z}
tan x
Integrals with tan x and sec x
Z
Z
2
sec x tan xdx −
sin x
dx
cos x
We make the substitutions:
u = tan x,
du
= sec2 x
dx
v = cos x,
dv
= − sin x
dx
Replacing in the integrals:
Z
du
u .
|{z}
dx
tan x |{z}
sec2 x
Integrals with tan x and sec x
Z
Z
2
sec x tan xdx −
sin x
dx
cos x
We make the substitutions:
u = tan x,
du
= sec2 x
dx
v = cos x,
dv
= − sin x
dx
Replacing in the integrals:
Z
du
u .
.dx
|{z}
dx
|{z}
tan x
sec2 x
Integrals with tan x and sec x
Z
Z
2
sec x tan xdx −
sin x
dx
cos x
We make the substitutions:
u = tan x,
du
= sec2 x
dx
v = cos x,
dv
= − sin x
dx
Replacing in the integrals:
Z
du
u .
.dx−
|{z}
dx
|{z}
tan x
sec2 x
Integrals with tan x and sec x
Z
Z
2
sec x tan xdx −
sin x
dx
cos x
We make the substitutions:
u = tan x,
du
= sec2 x
dx
v = cos x,
dv
= − sin x
dx
Replacing in the integrals:
Z
Z
du
1
u
.
.dx
−
|{z} dx
v
|{z}
tan x |{z}
sec2 x
cos x
Integrals with tan x and sec x
Z
2
Z
sec x tan xdx −
sin x
dx
cos x
We make the substitutions:
u = tan x,
du
= sec2 x
dx
v = cos x,
dv
= − sin x
dx
Replacing in the integrals:
Z
Z
1
du
dv
u .
.dx −
. −
|{z}
dx
v
dx
|{z}
tan x |{z}
| {z }
sec2 x
cos x
sin x
Integrals with tan x and sec x
Z
2
Z
sec x tan xdx −
sin x
dx
cos x
We make the substitutions:
u = tan x,
du
= sec2 x
dx
v = cos x,
dv
= − sin x
dx
Replacing in the integrals:
Z
Z
1
du
dv
.dx
u .
.dx −
. −
|{z}
dx
v
dx
|{z}
tan x |{z}
| {z }
sec2 x
cos x
sin x
Integrals with tan x and sec x
Z
2
Z
sec x tan xdx −
sin x
dx
cos x
We make the substitutions:
u = tan x,
du
= sec2 x
dx
v = cos x,
dv
= − sin x
dx
Replacing in the integrals:
Z
Z
1
du
dv
.dx
u .
.dx −
. −
|{z}
v
dx
dx
|{z}
tan x |{z}
| {z }
sec2 x
cos x
sin x
Integrals with tan x and sec x
Z
2
Z
sec x tan xdx −
sin x
dx
cos x
We make the substitutions:
u = tan x,
du
= sec2 x
dx
v = cos x,
dv
= − sin x
dx
Replacing in the integrals:
Z
Z
1
du
dv
.dx
u .
.dx −
. −
|{z}
v
dx
dx
|{z} | {z }
tan x |{z}
sec2 x
cos x
sin x
Integrals with tan x and sec x
Z
Z
2
sec x tan xdx −
sin x
dx
cos x
We make the substitutions:
u = tan x,
du
= sec2 x
dx
v = cos x,
dv
= − sin x
dx
Replacing in the integrals:
Z
Z
1
du
dv
.dx
u .
.dx −
. −
|{z}
v
dx
dx
|{z} | {z }
tan x |{z}
cos x
sec2 x
=
sin x
Integrals with tan x and sec x
Z
Z
2
sec x tan xdx −
sin x
dx
cos x
We make the substitutions:
u = tan x,
du
= sec2 x
dx
v = cos x,
dv
= − sin x
dx
Replacing in the integrals:
Z
Z
1
du
dv
.dx
u .
.dx −
. −
|{z}
v
dx
dx
|{z} | {z }
tan x |{z}
cos x
sec2 x
Z
=
sin x
Integrals with tan x and sec x
Z
Z
2
sec x tan xdx −
sin x
dx
cos x
We make the substitutions:
u = tan x,
du
= sec2 x
dx
v = cos x,
dv
= − sin x
dx
Replacing in the integrals:
Z
Z
1
du
dv
.dx
u .
.dx −
. −
|{z}
v
dx
dx
|{z} | {z }
tan x |{z}
cos x
sec2 x
Z
=
u
sin x
Integrals with tan x and sec x
Z
Z
2
sec x tan xdx −
sin x
dx
cos x
We make the substitutions:
u = tan x,
du
= sec2 x
dx
v = cos x,
dv
= − sin x
dx
Replacing in the integrals:
Z
Z
1
du
dv
.dx
u .
.dx −
. −
|{z}
v
dx
dx
|{z} | {z }
tan x |{z}
cos x
sec2 x
Z
=
udu
sin x
Integrals with tan x and sec x
Z
Z
2
sec x tan xdx −
sin x
dx
cos x
We make the substitutions:
u = tan x,
du
= sec2 x
dx
v = cos x,
dv
= − sin x
dx
Replacing in the integrals:
Z
Z
1
du
dv
.dx
u .
.dx −
. −
|{z}
v
dx
dx
|{z} | {z }
tan x |{z}
cos x
sec2 x
Z
=
udu+
sin x
Integrals with tan x and sec x
Z
Z
2
sec x tan xdx −
sin x
dx
cos x
We make the substitutions:
u = tan x,
du
= sec2 x
dx
v = cos x,
dv
= − sin x
dx
Replacing in the integrals:
Z
Z
1
du
dv
.dx
u .
.dx −
. −
|{z}
v
dx
dx
|{z} | {z }
tan x |{z}
cos x
sec2 x
Z
=
Z
udu +
sin x
Integrals with tan x and sec x
Z
Z
2
sin x
dx
cos x
sec x tan xdx −
We make the substitutions:
u = tan x,
du
= sec2 x
dx
v = cos x,
dv
= − sin x
dx
Replacing in the integrals:
Z
Z
1
du
dv
.dx
u .
.dx −
. −
|{z}
v
dx
dx
|{z} | {z }
tan x |{z}
cos x
sec2 x
Z
=
Z
udu +
dv
v
sin x
Integrals with tan x and sec x
Z
Z
udu +
dv
v
Integrals with tan x and sec x
Z
Z
udu +
We know how to solve this:
dv
v
Integrals with tan x and sec x
Z
Z
udu +
We know how to solve this:
=
u2
2
dv
v
Integrals with tan x and sec x
Z
Z
udu +
We know how to solve this:
=
u2
+
2
dv
v
Integrals with tan x and sec x
Z
Z
udu +
dv
v
We know how to solve this:
=
u2
+ ln v
2
Integrals with tan x and sec x
Z
Z
udu +
dv
v
We know how to solve this:
=
u2
+ ln v + C
2
Integrals with tan x and sec x
Z
Z
udu +
dv
v
We know how to solve this:
u2
+ ln v + C
2
And now, we substitute back:
=
Integrals with tan x and sec x
Z
Z
udu +
dv
v
We know how to solve this:
u2
+ ln v + C
2
And now, we substitute back:
=
u=
Integrals with tan x and sec x
Z
Z
udu +
dv
v
We know how to solve this:
u2
+ ln v + C
2
And now, we substitute back:
=
u = tan x
Integrals with tan x and sec x
Z
Z
udu +
dv
v
We know how to solve this:
u2
+ ln v + C
2
And now, we substitute back:
=
u = tan x,
Integrals with tan x and sec x
Z
Z
udu +
dv
v
We know how to solve this:
u2
+ ln v + C
2
And now, we substitute back:
=
u = tan x,
v = cos x
Integrals with tan x and sec x
Z
Z
udu +
dv
v
We know how to solve this:
u2
+ ln v + C
2
And now, we substitute back:
=
u = tan x,
So, our integral is:
v = cos x
Integrals with tan x and sec x
Z
Z
udu +
dv
v
We know how to solve this:
u2
+ ln v + C
2
And now, we substitute back:
=
u = tan x,
v = cos x
So, our integral is:
Z
tan3 xdx
Integrals with tan x and sec x
Z
Z
udu +
dv
v
We know how to solve this:
u2
+ ln v + C
2
And now, we substitute back:
=
u = tan x,
v = cos x
So, our integral is:
Z
tan3 xdx =
Integrals with tan x and sec x
Z
Z
udu +
dv
v
We know how to solve this:
u2
+ ln v + C
2
And now, we substitute back:
=
u = tan x,
v = cos x
So, our integral is:
Z
tan3 xdx =
tan2 x
2
Integrals with tan x and sec x
Z
Z
udu +
dv
v
We know how to solve this:
u2
+ ln v + C
2
And now, we substitute back:
=
u = tan x,
v = cos x
So, our integral is:
Z
tan3 xdx =
tan2 x
+
2
Integrals with tan x and sec x
Z
Z
udu +
dv
v
We know how to solve this:
u2
+ ln v + C
2
And now, we substitute back:
=
u = tan x,
So, our integral is:
Z
tan3 xdx =
v = cos x
tan2 x
+ ln cos x
2
Integrals with tan x and sec x
Z
Z
udu +
dv
v
We know how to solve this:
u2
+ ln v + C
2
And now, we substitute back:
=
u = tan x,
So, our integral is:
Z
tan3 xdx =
v = cos x
tan2 x
+ ln cos x + C
2
Integrals with tan x and sec x
Z
Z
udu +
dv
v
We know how to solve this:
u2
+ ln v + C
2
And now, we substitute back:
=
u = tan x,
v = cos x
So, our integral is:
Z
tan3 xdx =
tan2 x
+ ln cos x + C
2