KUIZ GUYYS OH YA AKU MUDIK :V SKARANG AJA BARU BRANGKATSoal:1)Apa Rumus Penjumlahan Sin Cos Tan?2) Sebutkan 3 Contoh Soal dan jawabannya Tentang Integral parsial![tex] \tt \infty yey \: bahagianya[/tex]​

1. KUIZ GUYYS OH YA AKU MUDIK :V SKARANG AJA BARU BRANGKATSoal:1)Apa Rumus Penjumlahan Sin Cos Tan?2) Sebutkan 3 Contoh Soal dan jawabannya Tentang Integral parsial![tex] \tt \infty yey \: bahagianya[/tex]​

Jawab:

Penjelasan dengan langkah-langkah:

[tex] \huge 1. [/tex]

sin (A ± B) = sin A cos B ± sin B cos A

cos (A ± B) = cos A cos B [tex] \frac{}{+} [/tex] sin A sin B

tan (A ± B) = [tex] \frac{ \tan A ± \tan B}{1 \frac{}{+} \tan A \tan B} [/tex]

[tex] \huge 2. [/tex]

[tex] \huge \text{Contoh soal 1} [/tex]

Tentukan hasil dari ∫ 2x (x + 7)⁹ dx

Gunakan integral parsial, misal kan :

∫ 2x (x + 7)⁹ dx = ∫ u dv

maka didapat :

u = 2x => du = 2 dx

dv = (x + 7)⁹ dx => v = ⅒(x + 7)¹⁰

hasil integral nya menjadi :

= 2x (⅒(x + 7)¹⁰) -∫ ⅒(x + 7)¹⁰ . 2 dx

= ⅕x (x + 7)¹⁰ -⅕ ∫ (x + 7)¹⁰ dx

= ⅕x (x + 7)¹⁰ -⅕ (1/11 (x + 7)¹¹)

= ⅕x (x + 7)¹⁰ -1/55 (x + 7)¹¹ + C

[tex] \huge \text{Contoh soal 2} [/tex]

Berapa hasil dari [tex] \displaystyle \int \sec (x) \text{dx} [/tex]

maka kita ubah bentuknya menjadi :

[tex] \\ \displaystyle \int \sec (x) \text{dx} = \displaystyle \int \sec (x) \cdot \frac{ \sec(x) + \tan(x) }{\sec(x) + \tan(x) } \text{dx} \\ = \displaystyle \int \frac{ \sec {}^{2} (x) + \sec(x) \tan(x) }{\sec(x) + \tan(x) } \text{dx} [/tex]

Gunakan integral substitusi, misalkan :

u = sec (x) + tan (x) [tex] \to \text{du} = \sec (x) \tan (x) + \sec ^2 (x) \text{dx} [/tex]

maka bentuknya diubah menjadi :

[tex] = \displaystyle \int \frac{ 1 }{u } \text{du} \\ = \ln(u) \\ \displaystyle \int \sec (x) \text{dx} = \ln | \sec(x) + \tan(x) | + C [/tex]

[tex] \huge \text{Contoh soal 3} [/tex]

Berapa hasil dari [tex] \displaystyle \int \sec ^3 (x) \text{dx} [/tex]

Nah, kita misalkan :

[tex] \displaystyle \int \sec (x) \cdot \sec ^2 (x) \text{dx} = \displaystyle \int u \: \text{dv} [/tex]

maka didapat :

u = sec (x) [tex] \to \text{du} = \sec (x) \cdot \tan (x) \text{dx} [/tex]

dv = sec²(x) dx [tex] \to v = \tan (x) [/tex]

gunakan integral parsial ;

[tex] \displaystyle \int u \: \text{dv} = u.v - \displaystyle \int v \: \text{du} [/tex]

[tex] \\ \displaystyle \int \sec (x) \cdot \sec ^2 (x) \text{dx} = ( \sec(x) )( \tan(x) ) - \displaystyle \int \tan(x) \cdot \sec(x) \tan(x) \text{dx} \\ [/tex]

[tex] \\ \displaystyle \int \sec (x) \cdot \sec ^2 (x) \text{dx} = ( \sec(x)) ( \tan(x) )- \displaystyle \int \tan(x) \cdot \sec(x) \tan(x) \text{dx} \\ \displaystyle \int \sec ^3 (x) \text{dx} = \sec(x) \tan(x) - \displaystyle \int \sec(x)( \sec {}^{2} (x) - 1) \text{dx} \\\displaystyle \int \sec {}^{3} (x) \text{dx} = \sec(x) \tan(x) - \displaystyle \int \sec {}^{3} (x)\text{dx}- \displaystyle \int \sec(x) \text{dx} \\ 2\displaystyle \int \sec {}^{3} (x) \text{dx} = \sec(x) \tan(x) - \ln( \sec(x) + \tan(x) ) \\ \displaystyle \int \sec {}^{3} (x) \text{dx} = \frac{1}{2} \sec(x) \tan(x) - \frac{1}{2} \ln | \sec(x) + \tan(x) | + C [/tex]

Jawaban:

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Penjelasan dengan langkah-langkah:

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