minta tolonh integral parsial ka

Berikut ini adalah pertanyaan dari ariem406 pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Minta tolonh integral parsial ka

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

1. Hasil dari $\displaystyle{f(x)=\int\limits {\sqrt{9+x^2}} \, dx }$ adalah $\displaystyle{\boldsymbol{f(x)=\frac{x\sqrt{9+x^2}}{2}+\frac{9}{2}ln\left | \frac{x+\sqrt{9+x^2}}{3} \right |+C} }$ .

2. Hasil dari $\displaystyle{f(x)=\int\limits^{\frac{\pi}{4}}_0 {\sqrt{9+x^2}} \, dx }$ adalah $\displaystyle{\boldsymbol{f(x)=\frac{\pi\sqrt{144+\pi^2}}{32}+\frac{9}{2}ln\left | \frac{\pi+\sqrt{144+\pi^2}}{12} \right |} }$ .

PEMBAHASAN

Salah satu metode untuk menyelesaikan integral adalah metode integral parsial, dimana :

$\displaystyle{\int\limits {u} \, dv=uv-\int\limits {v} \, du }$

DIKETAHUI

$\displaystyle{1.~f(x)=\int\limits {\sqrt{9+x^2}} \, dx }$

$\displaystyle{2.~f(x)=\int\limits^{\frac{\pi}{4}}_0 {\sqrt{9+x^2}} \, dx }$

DITANYA

Tentukan hasilnya.

PENYELESAIAN

Soal 1.

Misal :

$\displaystyle{tan\theta=\frac{x}{3}~\to~sec\theta=\frac{\sqrt{9+x^2}}{3}}$

$\displaystyle{3tan\theta=x}$

$\displaystyle{3sec^2\theta d\theta=dx}$

$\displaystyle{\int\limits {\sqrt{9+x^2}} \, dx=\int\limits {\sqrt{9+(3tan\theta)^2}} \, (3sec^2\theta d\theta) }$

$\displaystyle{\int\limits {\sqrt{9+x^2}} \, dx=3\int\limits {sec^2\theta\sqrt{9(1+tan^2\theta)}} \, d\theta }$

$\displaystyle{\int\limits {\sqrt{9+x^2}} \, dx=3\int\limits {sec^2\theta\sqrt{9sec^2\theta}} \, d\theta }$

$\displaystyle{\int\limits {\sqrt{9+x^2}} \, dx=9\int\limits {sec^3\theta} \, d\theta }$

$\displaystyle{\int\limits {\sqrt{9+x^2}} \, dx=9\int\limits {sec\theta sec^2\theta} \, d\theta }$

$-----------$

Mencari $\displaystyle{\int\limits {sec\theta sec^2\theta} \, d\theta }$ :

Misal :

$u=sec\theta~\to~du=sec\theta tan\theta d\theta$

$dv=sec^2\theta d\theta~\to~v=tan\theta$

$\displaystyle{\int\limits {sec\theta sec^2\theta} \, d\theta=uv-\int\limits {v} \, du }$

$\displaystyle{\int\limits {sec\theta sec^2\theta} \, d\theta=sec\theta tan\theta-\int\limits {tan\theta} \, (sec\theta tan\theta d\theta) }$

$\displaystyle{\int\limits {sec\theta sec^2\theta} \, d\theta=sec\theta tan\theta-\int\limits {sec\theta tan^2\theta} \, d\theta }$

$\displaystyle{\int\limits {sec\theta sec^2\theta} \, d\theta=sec\theta tan\theta-\int\limits {sec\theta(sec^2\theta-1)} \, d\theta }$

$\displaystyle{\int\limits {sec\theta sec^2\theta} \, d\theta=sec\theta tan\theta-\int\limits {sec\theta sec^2\theta} \, d\theta+\int\limits {sec\theta} \, d\theta }$

$\displaystyle{2\int\limits {sec\theta sec^2\theta} \, d\theta=sec\theta tan\theta+\int\limits {sec\theta} \, d\theta }$

$\displaystyle{2\int\limits {sec\theta sec^2\theta} \, d\theta=sec\theta tan\theta+ln|tan\theta+sec\theta|+C }$

$\displaystyle{\int\limits {sec\theta sec^2\theta} \, d\theta=\frac{1}{2}\left ( sec\theta tan\theta+ln|tan\theta+sec\theta| \right )+C }$

$-----------$

$\displaystyle{\int\limits {\sqrt{9+x^2}} \, dx=9\int\limits {sec\theta sec^2\theta} \, d\theta }$

$\displaystyle{\int\limits {\sqrt{9+x^2}} \, dx =\frac{9}{2}\left ( sec\theta tan\theta+ln|tan\theta+sec\theta| \right )+C }$

Substitusi kembali nilai tanθ dan secθ

$\displaystyle{\int\limits {\sqrt{9+x^2}} \, dx=\frac{9}{2}\left [ \frac{\sqrt{9+x^2}}{3} \frac{x}{3}+ln\left | \frac{\sqrt{9+x^2}}{3}+\frac{x}{3} \right | \right ]+C }$

$\displaystyle{\int\limits {\sqrt{9+x^2}} \, dx=\frac{9}{2}\left [ \frac{x\sqrt{9+x^2}}{9} +ln\left | \frac{x+\sqrt{9+x^2}}{3} \right | \right ]+C }$

$\displaystyle{\int\limits {\sqrt{9+x^2}} \, dx=\frac{x\sqrt{9+x^2}}{2}+\frac{9}{2}ln\left | \frac{x+\sqrt{9+x^2}}{3} \right |+C }$

Soal 2.

$\displaystyle{f(x)=\int\limits^{\frac{\pi}{4}}_0 {\sqrt{9+x^2}} \, dx }$

$\displaystyle{f(x)=\frac{x\sqrt{9+x^2}}{2}+\frac{9}{2}ln\left | \frac{x+\sqrt{9+x^2}}{3} \right |\Bigr|^{\frac{\pi}{4}}_{0} }$

$\displaystyle{f(x)=\frac{\frac{\pi}{4}\sqrt{9+\frac{\pi^2}{16}}}{2}+\frac{9}{2}ln\left | \frac{\frac{\pi}{4}+\sqrt{9+\frac{\pi^2}{16}}}{3} \right |-\frac{0\sqrt{9+0}}{2}+\frac{9}{2}ln\left | \frac{0+\sqrt{9+0}}{3} \right | }$

$\displaystyle{f(x)=\frac{\pi\sqrt{\frac{144+\pi^2}{16}}}{8}+\frac{9}{2}ln\left | \frac{\frac{\pi}{4}+\sqrt{\frac{144+\pi^2}{16}}}{3} \right |-0+\frac{9}{2}ln\left | 1 \right | }$