integral sec x + tan x
1. integral sec x + tan x
Penjelasan dengan langkah-langkah:
maaaf kak kalau salah... ..mm ..
2. Hasil dari integral sec x (tan x + sec x) dx adalah
Integral Fungsi Trigonometri.
∫ sec x (tan x + sec x) dx
= ∫ (sec x tan x + sec² x) dx
= sec x + tan x + C
= sec x + 1 / cos x + C
3. hasil integral sec x (tan x + sec x) dx
integr sec x (tanx + sec x) dx
= integr secx . tanx dx + integr sec^2 x dx
= sec x + tan x + c
4. (tan x + sec x)(tan x - sec x)=
semoga membantu.........
[tex] \tan(x) = \frac{ \sin(x) }{ \cos(x) } \\ \sec(x) = \frac{1}{ \cos(x) } [/tex]
[tex] (\tan \: x + \sec \: x )(\tan \: x - \sec \: x) \\ = ( \frac{ \sin(x) }{ \cos(x) } + \frac{1}{ \cos(x) } )( \frac{ \sin(x) }{ \cos(x) } - \frac{1}{ \cos(x) } ) \\ = ( \frac{ \sin(x) + 1 }{ { \cos \:}x } ) ( \frac{ \sin(x) - 1 }{ { \cos \:}x } ) \\ = ( \frac{ { \sin }^{2}x - 1 }{ \cos ^{2}x } ) \\ = \frac{ { \cos }^{2}x }{ { \cos }^{2}x } \\ = 1[/tex]
Semoga dapat dipahami
5. sec⁴x-sec² x=tan⁴ + tan² x
Soalnya:
Sec⁴x-sec²x=tan⁴ + tan² x
Jawaban:
1/cos⁴ - 1/cos²x = Sin⁴x/ cos²x + Sin²x/cos²x
=tan⁴ x + tan² x
Maaf kalo salah ya
6. Integral dari (tan x - sec x)^2 adalah? Sama caranya yaa?
↑=anggap lambang integral...
↑ (tanx - secx )² = ↑tan²x + sec²x - 2 secx tanx dx
= ↑2tan²x + 1 dx - ↑2secxtanx dx
= - 2 ln [cosx] + x - 2 secx
semoga membantu, sukses ya dek
jgn lupa jadiin jawaban terbaik lho
7. Integral(tan x-sec x)2 dx
Integral Fungsi Trigonometri.
∫ (tan x - sec x)² dx
= ∫ (tan² x - 2 tan x sec x + sec² x) dx
= ∫ [(sec² x - 1) - 2 tan x sec x + sec² x] dx
= tan x - x - 2 sec x + tan x + C
= 2 (tan x - sec x) - x + C
8. integral tan x sec^2 dx
∫ tan x(sec² x) dx
= ∫ sec x d(tan x.sec x)
= (1/2).sec² x + C
9. integral x sec² (1+2x²) tan (1+2x²) dx
Jawab:
set
u = 1 + 2x²
du = 4x dx
x dx = 1/4 du
∫x sec² (1+2x²) tan (1+2x²) dx
= 1/4 ∫ tan u sec² u d u
set
v = tan u
dv = sec² u du
= 1/4 ∫ tan u sec² u d u|
= 1/4 ∫ v dv
= 1/4. 1/2 . v² + c
= 1/8 v² + c
= 1/8 tan² u + c
= 1/8 tan² (1 + 2x² ) + c
10. 1. integral (cosec^2 x - sec x • tan x) dx2. integral ( sec^2 x + cosec ^2 x) dx3. integral ( tan x • sec x + cot x • cosec x) dx4. integral (4 sin x-2 cos x) dx
Jawaban:
gak tau aku mah belum belajar yang kayak begituan
11. Integral ( 4 sec x + 5 tan x) sec dx
∫(sec x)(4sec x + 5 tan x) dx
= ∫4sec² x dx + ∫5 sec x tan x dx
= 4 ∫dtan x + 5 ∫dsec x
= 4tan x + 5 sec x + C
12. integral 1 - tan kuadrat x bagi sec kuadrat x adalah
Jawab:
[tex]\int {\frac{1-tan^2x}{sec^2x}} \, dx=\frac{1}{2}sin2x+C[/tex]
Penjelasan dengan langkah-langkah:
INTEGRAL TAK TENTU
.
Diketahui :
[tex]\int {\frac{1-tan^2x}{sec^2x}} \, dx=[/tex]
.
Ditanya :
hasil integral fungsi tersebut
.
Penyelesaian :
[tex]\int {\frac{1-tan^2x}{sec^2x}} \, dx\\\\=\int {cos^2x(1-tan^2x)} \, dx\\\\=\int {cos^2x(1-(sec^2x-1))} \, dx\\\\=\int {cos^2x(2-sec^2x)} \, dx\\\\=\int {2cos^2x-cos^2xsec^2x} \, dx\\\\=\int {2cos^2x-1} \, dx\\\\=\int {cos2x} \, dx\\\\=\frac{1}{2}sin2x+C[/tex]
.
note : identitas trigonometri yang digunakan :
[tex]1+tan^2x=sec^2x\\\\secx=\frac{1}{cosx}\\\\cos2x=2cos^2x-1[/tex]
.
Pelajari Lebih Lanjut :
> integral fungsi trigonometri : https://brainly.co.id/tugas/28441698
> integral fungsi trigonometri : https://brainly.co.id/tugas/28222821
.
#sejutapohon
Mapel: Matematika
Kelas : 11
Bab : Integral Fungsi Aljabar
Kata Kunci: integral, tak, tentu, fungsi, fungsi, trigonometri
Kode Kategorisasi: 11.2.10
13. integral 4x^2 sec x^3 tan x^ 3 dx
Perhitungan Terlampir
14. integral 3 sec x tan x dx
Materi integral
int (3 sec x tan x) dx
= 3 int (sec x tan x) dx
= 3 sec x + C
15. tentukan hasil pengintegralan dari 3 sec x tan x dx
integral
∫3 sec x tan x dx
misal u = sec x
du = sec x tan x dx
∫3 sec x tan x dx = ∫ 3 du = 3u+ c = 3 sec x + c
16. hasil integral sec x (tan x + sec x) dx= dengan caranya ya
int sec x (tan x + sec x) dx
=> int (sec x tan x + sec^2 x) dx
=> sec x + tan x + C
17. integral tan pangkat 3 sec pangkat -3 x dx
Jawab:
Penjelasan dengan langkah-langkah:
[tex]\int {tan^3x.sec^{-3}x} \, dx\\\\=\int {\frac{sin^3x}{cos^3x}.cos^3x} \, dx\\\\=\int {sinx.sin^2x} \, dx\\\\=\int {sinx(1-cos^2x)} \, dx\\\\=\int {sinx-sinx.cos^2x} \, dx\\\\=\int {sinx} \, dx-\int {sinx.cos^2x} \, dx\\\\=-cosx-\int {sinx.cos^2x} \, dx~~~~~~~~~~~...misal~u=cosx~\to~du=-sinxdx\\\\=-cosx-\int {sinx.u^2} \, \frac{du}{-sinx}\\\\=-cosx+\int {u^2} \, du\\\\=-cosx+\frac{1}{3}u^3+C\\\\=-cosx+\frac{1}{3}cos^3x+C[/tex]
#sejutapohon
Mapel: Matematika
Kelas : 11
Bab : Integral
Kata Kunci: integral, trigonometri, substitusi
Kode Kategorisasi: 11.2.10
18. integral (cosec^2 x - sec x • tan x) dx
integral trigonometri
.
∫ (cosec² x - sec x tan x ) dx
= ∫ (cosec² x) dx - ∫ (sec x tan x ) dx
= cot x - sec x + c
19. tolong di jawab integral sec x tan x per 9+4 sec pangkat 2 x
Materi : Kalkulus Integral
Mungkin begini y :
[tex]\int{\frac{\sec{x}.\tan{x}}{9+4\sec^2{x}}\,dx}[/tex]
Misalkan :
u = sec x
du = sec x.tan x dx
Maka :
[tex]\int{\frac{\sec{x}.\tan{x}}{9+4\sec^2{x}}\,dx}=\int{\frac{du}{9+4u^2}}[/tex]
Substitusi :
u = [tex]\frac{3}{2}\tan{t}[/tex]
[tex]\tan{t}=\frac{2}{3}u\\t=\arctan{\frac{2}{3}u}[/tex]
du = [tex]\frac{3}{2}\sec^2{t}[/tex] dt
Maka :
[tex]\int{\frac{du}{9+4u^2}}=\int{\frac{\frac{3}{2}\sec^2{t}\,dt}{9(1+tan^2{t})}}=\frac{3}{2}\int{\frac{\sec^2{t}\,dt}{9\sec^2{t}}}\\\int{\frac{du}{9+4u^2}}=\frac{1}{6}\int{\,dt}=\frac{1}{6}t+c[/tex]
Jadi :
[tex]\int{\frac{du}{9+4u^2}}=\frac{1}{6}t+c=\frac{1}{6}\arctan{\frac{2}{3}u}+c[/tex]
Dengan mengembalikan u dalam x, hasil integralnya diperoleh, yaitu :
[tex]\int{\frac{\sec{x}.\tan{x}}{9+4\sec^2{x}}\,dx}=\frac{1}{6}\arctan{\left(\frac{2}{3}\sec{x}\right)}+c[/tex]
Semoga membantu.
20. integral sec x²/tan x dx ?
int sec²x/tanx dx
misal,
u = tanx
du = sec²x dx
so,
= int du/u
= ln(u) +c
= ln(tanx) + c
21. tentukanlah integral (sec^2x - tan x sec x)dx
integral [(sec^2x)- (tan x sec x)]dx
=tan x - sec x + c
22. hasil integral tak tentu fungsi trigonometri dari Integral (csc x cot x - sec x tan x) dx = .... # Asal jawab dihapus
semoga bermanfaat ya
23. hasil dari integral sec x (tan x + sec x) dx adalah
∫ sec x (tan x + sec x) dx
= ∫ sec x . tan x dx + ∫ (sec x)^2 dx
= sec x + tan x + C
CMIIW
24. integral (sec x - tan x)² dx
integral (sec x - tan x)² dx=
integral (sec² x - 2sec x tan x+tan² x) dx=
integral (sec² x - 2(sin x)/cos²x+sec² x-1) dx
integral (2sec² x - 2(sin x)/cos²x - 1) dx=
2tanx -x - integral ( 2(sin x)/cos²x )dx=
2tanx -x + integral (2/cos²x )d(cosx)=
2tanx -x -2/cosx +c=
2tanx -x -2secx+c
25. (tan x + sec x)(tan x - sec x)=
Uraian lihat foto ya
26. integral parsial [tex] \tan^{4} (x) \sec ^{2} (x) dx[/tex]
Jawab:
Penjelasan dengan langkah-langkah:
t = tan(x), dt = sec²(x).dx
[tex]\displaystyle \int {\tan^{4}(x).\sec^{2}(x)} \, dx = \int {t^{4}}\,dt[/tex]
[tex]\displaystyle \int {\tan^{4}(x).\sec^{2}(x)} \,dx = \frac{t^{5}}{5} + C = \frac{\tan^{5}(x)}{5}+C[/tex]
27. Hitunglah integral dari tan x sec² x
int tan x sec² x dx =
misal u= tanx
int tan x d(tanx )=½tan²x+c
28. integral dari (tan x . sec x + cot x . cosec x) dx =
[tex]\displaystyle \int {\tan x\sec x+\cot x\csc x} \, dx =\sec x-\csc x+C[/tex]
29. 1.) Integral cos x ( sec x + 1 ) dx 2.) Integral cos x ( Tan x + sin x ) dx
[tex]\displaystyle1.\int\;\cos(x)(\sec(x)+1)\; \, dx = \int\;(\cos(x)+1)\; \, dx \\\\ \boxed{\int\;\cos(x)(\sec(x)+1)\; \, dx = \sin(x) + x + C}\\\\ 2. \int\;\cos(x)(\tan(x)+\sin(x))\; \, dx = \int\;(\sin(x)+\sin(x)\cos(x))\; \, dx\\\\ \int\;\cos(x)(\tan(x)+\sin(x))\; \, dx = \int\;\sin(x)(1+\cos(x))\; \, dx\\\\ t = \cos(x), dt = -\sin(x)\;dx\\\\\int\;\cos(x)(\tan(x)+\sin(x))\; \, dx = -\int\;(1+t)\; \, dt\\\\ \int\;\cos(x)(\tan(x)+\sin(x))\; \, dx = -\frac{(t+1)^2}{2}[/tex]
[tex]\displaystyle \boxed{\int\;\cos(x)(\tan(x)+\sin(x))\; \, dx = -\frac{(\cos(x)+1)^2}{2}}[/tex]
30. integral dari tan x sec² x dx
Int tan x sec² x dx
= Int tan x dtan x
= (1/2)tan² x + C
31. integral 2x sec x^2 tan x^2 dx
misal x^2= u
du = 2x dx
∫ (2x). sex (x²) tan (x²) dx = ∫ sec u tan u du = sec u + c
= sec (x²) + C
32. (sec x - tan x) (sec x. tan x)
Jadikan yang terbaik ya...
(sec x - tan x) (sec x. tan x)
= [(1 / cos x) - (sin x / cos x)] [(1 / cos x) (sin x / cos x)]
= [(1 - sin x) / cos x] [sin x / cos² x]
= (sin x - sin² x) / cos³ x
~sen
33. integral tan pangkat 3 sec pangkat -3 x dx
Jawab:
[tex]\displaystyle\int\tan^3x\sec^{-3}x\,dx=\int\frac{\tan^3x}{\sec^3x}\,dx\\\int\tan^3x\sec^{-3}x\,dx=\int\frac{\displaystyle\frac{\sin^3x}{\cos^3x}}{\displaystyle\frac1{\cos^3x}}\,dx\\\int\tan^3x\sec^{-3}x\,dx=\int\sin^3x\,dx\\\int\tan^3x\sec^{-3}x\,dx=\int\sin x\sin^2x\,dx\\\int\tan^3x\sec^{-3}x\,dx=\int\sin x(1-\cos^2x)\,dx\\\int\tan^3x\sec^{-3}x\,dx=\int\sin x\,dx-\int\cos^2x\sin x\,dx\\\int\tan^3x\sec^{-3}x\,dx=\int\sin x\,dx-\int-\cos^2x\,d(\cos x)\\\int\tan^3x\sec^{-3}x\,dx=-\cos x+\frac{1}{2+1}\cos^{2+1}x+C\\\int\tan^3x\sec^{-3}x\,dx=\frac{1}3\cos^3x-\cos x+C[/tex]
Beberapa konsep yang dipakai:
[tex]\displaystyle \triangleright~\tan x=\frac{\sin x}{\cos x}\\\triangleright~\sec x=\frac1{\cos x}\\\triangleright~\int\sin x=-\cos x+C\\\triangleright~\int ax^n\,dx=\frac a{n+1}x^{n+1}+C~;~n\neq-1[/tex]
34. integral sec (x) tan (x) dx
int sec (x) tan (x) dx =
sec (x)
35. Berapakah hasil dari: 1. Integral tan (4x-2) sec dx 2. Integral (tan²x+4) dx 3. Integral (sin x-cos x)² dx Bantu ya
Integral
∫(tan² x + 4) dx
= ∫(sec² x - 1 + 4) dx
= ∫(sec² x + 3) dx
= tan x + 3x + C
•
∫(sin x - cos x)² dx
= ∫(1 - sin 2x) dx
= x + 1/2 ∫dcos 2x
= x + 1/2 cos 2x + C
Jawab:
`1. 6
Penjelasan dengan langkah-langkah:
36. (Tan x + sec X) (tan x - sec x)
Bab Trigonometri
Matematika SMA Kelas X
(tan x + sec x) (tan x - sec)
= tan² x - sec² x
= - (sec² x - tan² x)
= - 1
37. integral sec^2 x per tan^3 x?
[tex]sec(x)=\frac{1}{cos(x)}\\ cotan(x)=\frac{cos(x)}{sin(x)}\\ \int {\frac{sec^2(x)}{tan^3(x)}}dx=\int {sec^2(x)(\frac{1}{tan^3(x)})}dx\\ =\int {\frac{1}{cos^2(x)}(cotan^3(x))}dx\\ =\int {\frac{1}{cos^2(x)}(\frac{cos^3(x)}{sin^3(x)})}dx\\ =\int{\frac{cos(x)}{sin^3(x)}}dx\\ selanjutnya,\ bisa\ menggunakan\ integral\ substitusi\\ misal:\\ a=sin(x)\\ lalu, turunkan\ kedua\ ruas,\ sehingga\\ da=cos(x)dx \longrightarrow dx=\frac{da}{cos(x)}\\ maka :\\[/tex][tex]\int{\frac{cos(x)}{sin^3(x)}}dx=\int{\frac{cos(x)}{a^3}}(\frac{da}{cos(x)})\\ =\int\frac{1}{a^3}da=\frac{1}{-2a^2}+C=\frac{1}{-2sin^2(x)}+C[/tex]
38. Integral sec (x) × tan (x) dx
penyelesaian terlampir ya.
39. 1. Integral [tan^5 x sec^6 x] dx. 2. Integral [tan^2 x sec x] dx.
Jawab:
1. [tex]\frac{1}{10} tan^{10}x + \frac{1}{4} tan^{8}x + \frac{1}{6} tan^{6}x + c[/tex]
2. [tex]\frac{sec\ x\ tan\ x\ - ln(sec\ x\ +\ tan\ x)}{2} + c[/tex]
Penjelasan dengan langkah-langkah:
1.
[tex]\int tan^{5} x\sec^{6}x\, dx\\\int{tan^{5} x (tan^{2}x + 1)^{2} } \, d(tan x)\\\int{tan^{5} x (tan^{4}x + 2 tan^{2}x + 1) } \, d(tan x)\\\int{tan^{9}x +2 \tan^{7}x + tan^{5}x } \, d(tan x)\\\frac{1}{10} tan^{10}x + \frac{1}{4} tan^{8}x + \frac{1}{6} tan^{6}x + c[/tex]
2.
[tex]\int tan^{2}x \sec x\, dx = \int sec^{3}x\ dx - \int sec\ x\ dx[/tex]
Gunakan integral partial pada integral sec³ x:
[tex]\int u \, dv = uv - \int v \, du[/tex]
Di mana:
[tex]u = sec\ x, dv = sec^{2}x\ dx\\du = sec\ x\ tan\ x\ dx, v = tan\ x\\[/tex]
Masukkan ke rumus di atas:
[tex]\int\sec^{3}x\ dx = sec\ x\tan\ x - \int\tan^{2}x\ sec\ x\ dx[/tex]
Soal di atas bisa ditulis menjadi:
[tex]\int tan^{2}x \sec x\, dx\\\int tan^{2}x \sec x\, dx = sec\ x\tan\ x - \int\tan^{2}x\ sec\ x\ dx - \int sec\ x\ dx\\2 \int tan^{2}x \sec x\, dx = sec\ x\ tan\ x - \int sec\ x\ dx\\\int tan^{2}x \sec x\, dx = \frac{sec\ x\tan\ x\ - \int sec\ x\ dx}{2}\\\int tan^{2}x \sec x\, dx = \frac{sec\ x\tan\ x\ - ln(sec\ x\ +\ tan\ x)}{2}\\[/tex]
40. (tan x + sec x) (tan x - sec x)
(tan x + sec x)(tan x - sec x) = tan2 x - sec2 x = -1
Pembahasan.
Jika diketahui ( a + b ) ( a - b ) maka sama saja dengan a² - b² .
Mari kita terapkan
( tan x - sec x ) ( tan x + sec x )
tan² x - sec² x
Kita pakai salah satu rumus identitas trigonometri dengan sedikit modifikasi.
1 + tan² x = sec² x
1 = sec² x - tan² x
_____________ × -1
-1 = -sec² x + tan² x
-1 = tan² x - sec² x
Maka
tan² x - sec² x = -1
