Tag: Integral

$\begin{aligned} & \displaystyle \int \dfrac{(3x^2 + 2)^2\sqrt{x}}{5\sqrt[3]{x^2}}~\text{d}x \\ & = \displaystyle \int \dfrac{9x^4 + 12x^2 + 4}{5x^{\frac{1}{6}}}~\text{d}x \\ & = \dfrac{9}{5}\int x^{\frac{23}{6}}~dx + \dfrac{12}{5}\int x^{\frac{11}{6}}~dx + \dfrac{4}{5}\int x^{-\frac{1}{6}}~\text{d}x \\ & = \dfrac{9}{5} \times \dfrac{6}{29} x^{\frac{29}{6}} + \dfrac{12}{5} \times \dfrac{6}{17} x^{\frac{17}{6}} + \dfrac{4}{5} \times \dfrac{6}{5} x^{\frac{5}{6}} \\ & = \boxed{\dfrac{54}{145}x^4\sqrt[6]{x^5} + \dfrac{72}{85}x^2\sqrt[6]{x^5} + \dfrac{24}{25}\sqrt[6]{x^5}} \end{aligned}$

$\displaystyle \int \dfrac{2}{3} \cos^4 x~\text{d}x = \dfrac{2}{3} \int \cos^4 x~\text{d}x$
Gunakan rumus reduksi berikut.
$\int \cos^n x~\text{d}x = \dfrac{n-1}{n} \displaystyle \int \cos^{n-2} x~\text{d}x + \dfrac{\cos^{n-1} x \sin x}{n}$
Untuk $n = 4$, diperoleh
$\dfrac{2}{3} \left(\dfrac{3}{4} \displaystyle \int \cos^2 x~\text{d}x + \dfrac{\cos^3 x \sin x}{4}\right)$
$= \dfrac{\cos x \sin x}{4} + \dfrac{x}{4}  + \dfrac{\cos^3 x \sin x}{6}$
$=\boxed{\dfrac{\sin 4x + 8 \sin 2x + 12x}{48} + C}$

Misal $u = 2y^3 + 3y^2 + 6y$ sehingga $\text{d}u = 6(y^2 + y + 1)~\text{d}y.$ Perhatikan bahwa integral di atas dapat ditulis juga menjadi
$\dfrac{5}{6}\displaystyle \int \dfrac{6(y^2 + y+1)}{\sqrt{2y^3 + 3y^2 + 6y}}~\text{d}y.$
Substitusikan $u$ dan $\text{d}u.$
$ \dfrac{5}{6} \displaystyle \int \dfrac{du}{u^{\frac{1}{2}}}$
$=\dfrac{5}{6} \times 2 \times \sqrt{u} + C$
$= \boxed{\dfrac{5}{3}\sqrt{2y^3 + 3y^2 + 6y} + C}$

Misal $u = y^3- 3y$ sehingga $du = 3y^2- 3 = 3(y^2- 1)~\text{d}y.$ Perhatikan bahwa integral di atas dapat ditulis juga menjadi
$ \dfrac{4}{3} \displaystyle \int \dfrac{3(y^2- 1)}{(y^3- 3y)^2}~\text{d}y$
Substitusikan $u$ dan $\text{d}u$,
$ \dfrac{4}{3} \displaystyle \int \dfrac{du}{u^2}$
$ = \dfrac{4}{3} \times (-1) \times \dfrac{1}{u} + C$
$ = \boxed{-\dfrac{4}{3(y^3- 3y)} + C}$

$\displaystyle\int_{-2}^{6} (3x^2 + 5x)~\text{d}x$
$ =\left[x^3 + \dfrac{5}{2}x^2\right]_{-2}^{6}$
$ = \left(6^3 + \dfrac{5}{2}(6)^2\right)- \left((-2)^3 + \dfrac{5}{2}(-2)^2\right)$
$ = 216 + 90 + 8- 10 = \boxed{304}$

Misalkan $u = \sqrt[3]{x^2}- 5$ sehingga $\text{d}u = \dfrac{2}{3\sqrt[3]{x}}~\text{d}x.$
Bentuk integral di atas dapat ditulis menjadi bentuk
$ \dfrac{3}{2} \displaystyle\int_{1}^{8} \dfrac{2(\sqrt[3]{x^2}- 5)^4}{3\sqrt[3]{x}}~\text{d}x$
Substitusi $u$ dan $\text{d}u.$
$ \dfrac{3}{2}\displaystyle\int_{-4}^{-1} u^4~\text{d}u = \dfrac{3}{2} \times \dfrac{1}{5} \left[u^5\right]_{-4}^{-1}$
$ = \dfrac{3}{10} \times ((-1)^5- (-4)^5) = \dfrac{3}{10} \times 1.023$
$= \boxed{\dfrac{3.069}{10}}$

Misalkan $u = \sin \theta$ sehingga $\text{d}u = \cos \theta~\text{d}\theta.$ Dengan demikian, integral di atas dapat ditulis menjadi
$\displaystyle \int_{0}^{\frac{1}{2}\sqrt{3}} u^5~\text{d}u$
$= \left[\dfrac{1}{6}u^6\right]_{0}^{\frac{1}{2}\sqrt{3}} = \dfrac{1}{6}\left(\dfrac{1}{2}\sqrt{3}\right)^6- 0$
$ = \dfrac{27}{384} = \dfrac{9}{128}$
Jadi, nilai dari $\int_{0}^{\frac{\pi}{3}} \sin^5 \theta \cos \theta~\text{d}\theta$ adalah $\boxed{\dfrac{9}{128}}$.

Misalkan $u = x- 5$ sehingga $\text{d}u = \text{d}x.$ Di lain sisi, permisalan itu ekuivalen dengan $x = u + 5$ sehingga
$\displaystyle \int \dfrac{5x}{(x- 5)^4}~\text{d}x$
$=\displaystyle \int \dfrac{5(u+5)}{u^4}~\text{d}u$
$=\displaystyle \int \dfrac{5u + 25}{u^4}~\text{d}u$
$= \displaystyle \int \dfrac{5}{u^3}~du + \int \dfrac{25}{u^4}~\text{d}u$
$=-\dfrac{5}{2u^2}-\dfrac{25}{3u^3}$
$=-\dfrac{5}{2(x-5)^2}-\dfrac{25}{3(x-5)^3}$
(samakan penyebutnya)
$= \dfrac{-15(x-5)- 50}{6(x-5)^3} = \boxed{\dfrac{-15x + 25}{6(x-5)^3}}$