Integral dari $$$\sqrt{64 - x^{2}}$$$

Kalkulator akan menemukan integral/antiturunan dari $$$\sqrt{64 - x^{2}}$$$, dengan menampilkan langkah-langkah.

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Masukan Anda

Temukan $$$\int \sqrt{64 - x^{2}}\, dx$$$.

Solusi

Misalkan $$$x=8 \sin{\left(u \right)}$$$.

Maka $$$dx=\left(8 \sin{\left(u \right)}\right)^{\prime }du = 8 \cos{\left(u \right)} du$$$ (langkah-langkah dapat dilihat »).

Selain itu, berlaku $$$u=\operatorname{asin}{\left(\frac{x}{8} \right)}$$$.

Integran menjadi

$$$\sqrt{64 - x^{2}} = \sqrt{64 - 64 \sin^{2}{\left( u \right)}}$$$

Gunakan identitas $$$1 - \sin^{2}{\left( u \right)} = \cos^{2}{\left( u \right)}$$$:

$$$\sqrt{64 - 64 \sin^{2}{\left( u \right)}}=8 \sqrt{1 - \sin^{2}{\left( u \right)}}=8 \sqrt{\cos^{2}{\left( u \right)}}$$$

Dengan asumsi bahwa $$$\cos{\left( u \right)} \ge 0$$$, diperoleh sebagai berikut:

$$$8 \sqrt{\cos^{2}{\left( u \right)}} = 8 \cos{\left( u \right)}$$$

Oleh karena itu,

$${\color{red}{\int{\sqrt{64 - x^{2}} d x}}} = {\color{red}{\int{64 \cos^{2}{\left(u \right)} d u}}}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=64$$$ dan $$$f{\left(u \right)} = \cos^{2}{\left(u \right)}$$$:

$${\color{red}{\int{64 \cos^{2}{\left(u \right)} d u}}} = {\color{red}{\left(64 \int{\cos^{2}{\left(u \right)} d u}\right)}}$$

Terapkan rumus reduksi pangkat $$$\cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$ dengan $$$\alpha= u $$$:

$$64 {\color{red}{\int{\cos^{2}{\left(u \right)} d u}}} = 64 {\color{red}{\int{\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)d u}}}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=\frac{1}{2}$$$ dan $$$f{\left(u \right)} = \cos{\left(2 u \right)} + 1$$$:

$$64 {\color{red}{\int{\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)d u}}} = 64 {\color{red}{\left(\frac{\int{\left(\cos{\left(2 u \right)} + 1\right)d u}}{2}\right)}}$$

Integralkan suku demi suku:

$$32 {\color{red}{\int{\left(\cos{\left(2 u \right)} + 1\right)d u}}} = 32 {\color{red}{\left(\int{1 d u} + \int{\cos{\left(2 u \right)} d u}\right)}}$$

Terapkan aturan konstanta $$$\int c\, du = c u$$$ dengan $$$c=1$$$:

$$32 \int{\cos{\left(2 u \right)} d u} + 32 {\color{red}{\int{1 d u}}} = 32 \int{\cos{\left(2 u \right)} d u} + 32 {\color{red}{u}}$$

Misalkan $$$v=2 u$$$.

Kemudian $$$dv=\left(2 u\right)^{\prime }du = 2 du$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$du = \frac{dv}{2}$$$.

Integralnya menjadi

$$32 u + 32 {\color{red}{\int{\cos{\left(2 u \right)} d u}}} = 32 u + 32 {\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ dengan $$$c=\frac{1}{2}$$$ dan $$$f{\left(v \right)} = \cos{\left(v \right)}$$$:

$$32 u + 32 {\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}} = 32 u + 32 {\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{2}\right)}}$$

Integral dari kosinus adalah $$$\int{\cos{\left(v \right)} d v} = \sin{\left(v \right)}$$$:

$$32 u + 16 {\color{red}{\int{\cos{\left(v \right)} d v}}} = 32 u + 16 {\color{red}{\sin{\left(v \right)}}}$$

Ingat bahwa $$$v=2 u$$$:

$$32 u + 16 \sin{\left({\color{red}{v}} \right)} = 32 u + 16 \sin{\left({\color{red}{\left(2 u\right)}} \right)}$$

Ingat bahwa $$$u=\operatorname{asin}{\left(\frac{x}{8} \right)}$$$:

$$16 \sin{\left(2 {\color{red}{u}} \right)} + 32 {\color{red}{u}} = 16 \sin{\left(2 {\color{red}{\operatorname{asin}{\left(\frac{x}{8} \right)}}} \right)} + 32 {\color{red}{\operatorname{asin}{\left(\frac{x}{8} \right)}}}$$

Oleh karena itu,

$$\int{\sqrt{64 - x^{2}} d x} = 16 \sin{\left(2 \operatorname{asin}{\left(\frac{x}{8} \right)} \right)} + 32 \operatorname{asin}{\left(\frac{x}{8} \right)}$$

Dengan menggunakan rumus $$$\sin{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\sin{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\cos{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 1 - 2 \alpha^{2}$$$, $$$\cos{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$, $$$\sinh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha^{2} + 1}$$$, $$$\sinh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha - 1} \sqrt{\alpha + 1}$$$, $$$\cosh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} + 1$$$, $$$\cosh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$, sederhanakan ekspresi:

$$\int{\sqrt{64 - x^{2}} d x} = 4 x \sqrt{1 - \frac{x^{2}}{64}} + 32 \operatorname{asin}{\left(\frac{x}{8} \right)}$$

Sederhanakan lebih lanjut:

$$\int{\sqrt{64 - x^{2}} d x} = \frac{x \sqrt{64 - x^{2}}}{2} + 32 \operatorname{asin}{\left(\frac{x}{8} \right)}$$

Tambahkan konstanta integrasi:

$$\int{\sqrt{64 - x^{2}} d x} = \frac{x \sqrt{64 - x^{2}}}{2} + 32 \operatorname{asin}{\left(\frac{x}{8} \right)}+C$$

Jawaban

$$$\int \sqrt{64 - x^{2}}\, dx = \left(\frac{x \sqrt{64 - x^{2}}}{2} + 32 \operatorname{asin}{\left(\frac{x}{8} \right)}\right) + C$$$A