∫cos(ln x)dx=dan∫(2/(x.ln x2))dx=.... Question from @hanamutiadewi

acim 1) parsial 2 kali
int cos(ln(x)) dx
misal u = cos(ln(x)) ----> du = -1/x sin(ln(x)) dx
dv = dx -----> v = x
formula :
int u dv = uv - int v du
so, int cos(ln(x)) dx
= x cos(ln(x)) + int sin(ln(x)) dx

untuk int sin(ln(x)) dx, partialkan lagi ....
u = sin(ln(x)) ------> du = 1/x cos(ln(x)) dx
dv = dx -----> v = x

so,
int cos(ln(x)) dx = x cos(ln(x)) + x sin(ln(x)) - int cos(ln(x)) dx
2 int cos(ln(x)) dx = x cos(ln(x)) + x sin(ln(x)) + C
int cos(ln(x)) dx = (x cos(ln(x)) + x sin(ln(x))) / 2 + C

2 votes Thanks 1

Takamori37 Atas.
Dengan integral parsial berulang:
$\int \cos(\ln x)\, dx \\ U=\cos(\ln x)~~~~~~dU=\frac{-\sin(\ln x)}{x}\, dx \\ dV=dx~~~~~~~~~~~~~V=x$
Maka:
$\int \cos(\ln x)\, dx=x\cos(\ln x)-\int x\times (-\frac{\sin(\ln x)}{x})\, dx \\ \int \cos(\ln x)\, dx=x\cos(\ln x)+\int\sin(\ln x)\, dx$
Untuk pengintegralannya:
$\int \sin(\ln x)\, dx \\ U=\sin(\ln x)~~~~~~dU=\frac{\cos(\ln x)}{x}\, dx \\ dV=dx~~~~~~~~~~~~~V=x$
Maka:
$\int \sin(\ln x)\, dx=x\sin(\ln x)-\int x\times \frac{\cos(\ln x)}{x}\, dx \\ \int \sin(\ln x)\, dx=x\sin(\ln x)-\int\cos(\ln x)\, dx$
Sehingga:
$\int \cos(\ln x)\, dx=x\cos(\ln x)+\int\sin(\ln x)\, dx \\ \int \cos(\ln x)\, dx=x\cos(\ln x)+(x\sin(\ln x)-\int\cos(\ln x)\, dx) \\ \int \cos(\ln x)\, dx=x\cos(\ln x)+x\sin(\ln x)-\int\cos(\ln x)\, dx \\ 2\int \cos(\ln x)\, dx=x(\cos(\ln x)+\sin(\ln x)) \\ \int \cos(\ln x)\, dx=\frac{x}{2}(\cos(\ln x)+\sin(\ln x))+C$

Bagian bawah:
$\displaystyle \int \frac{2}{x\ln x^2}\, dx \\\\ U=\ln x^2 \\\\ dU=\frac{2}{x}\, dx \\\\ \text{Menjadi:} \\\\ \int\frac{1}{U}\, dU =\ln|U|+C \\\\ \ln(\ln x^2)+C$

3 votes Thanks 1

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