Integral dari $$$- \tan^{2}{\left(x \right)}$$$

Kalkulator akan menemukan integral/antiturunan dari $$$- \tan^{2}{\left(x \right)}$$$, dengan menampilkan langkah-langkah.

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Silakan tulis tanpa diferensial seperti $$$dx$$$, $$$dy$$$, dll.
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Masukan Anda

Temukan $$$\int \left(- \tan^{2}{\left(x \right)}\right)\, dx$$$.

Solusi

Terapkan aturan pengali konstanta $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ dengan $$$c=-1$$$ dan $$$f{\left(x \right)} = \tan^{2}{\left(x \right)}$$$:

$${\color{red}{\int{\left(- \tan^{2}{\left(x \right)}\right)d x}}} = {\color{red}{\left(- \int{\tan^{2}{\left(x \right)} d x}\right)}}$$

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

Kemudian $$$x=\operatorname{atan}{\left(u \right)}$$$ dan $$$dx=\left(\operatorname{atan}{\left(u \right)}\right)^{\prime }du = \frac{du}{u^{2} + 1}$$$ (langkah-langkahnya dapat dilihat »).

Integralnya menjadi

$$- {\color{red}{\int{\tan^{2}{\left(x \right)} d x}}} = - {\color{red}{\int{\frac{u^{2}}{u^{2} + 1} d u}}}$$

Tulis ulang dan pisahkan pecahannya:

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

Integralkan suku demi suku:

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

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

$$\int{\frac{1}{u^{2} + 1} d u} - {\color{red}{\int{1 d u}}} = \int{\frac{1}{u^{2} + 1} d u} - {\color{red}{u}}$$

Integral dari $$$\frac{1}{u^{2} + 1}$$$ adalah $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$:

$$- u + {\color{red}{\int{\frac{1}{u^{2} + 1} d u}}} = - u + {\color{red}{\operatorname{atan}{\left(u \right)}}}$$

Ingat bahwa $$$u=\tan{\left(x \right)}$$$:

$$\operatorname{atan}{\left({\color{red}{u}} \right)} - {\color{red}{u}} = \operatorname{atan}{\left({\color{red}{\tan{\left(x \right)}}} \right)} - {\color{red}{\tan{\left(x \right)}}}$$

Oleh karena itu,

$$\int{\left(- \tan^{2}{\left(x \right)}\right)d x} = - \tan{\left(x \right)} + \operatorname{atan}{\left(\tan{\left(x \right)} \right)}$$

Sederhanakan:

$$\int{\left(- \tan^{2}{\left(x \right)}\right)d x} = x - \tan{\left(x \right)}$$

Tambahkan konstanta integrasi:

$$\int{\left(- \tan^{2}{\left(x \right)}\right)d x} = x - \tan{\left(x \right)}+C$$

Jawaban

$$$\int \left(- \tan^{2}{\left(x \right)}\right)\, dx = \left(x - \tan{\left(x \right)}\right) + C$$$A


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