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人気のある微分積分 >

integral of 1/(cot^2(8x))sec^4(8x)

解

\int \frac{1}{cot ^{2} ( 8 x )} sec^{4} (8 x) d x

解

\frac{1}{8} (\frac{tan ^{3} ( 8 x )}{3} + \frac{tan ^{5} ( 8 x )}{5}) + C

解答ステップ

\int \frac{1}{cot ^{2} ( 8 x )} sec^{4} (8 x) d x

簡素化

\frac{1}{cot ^{2} ( 8 x )} sec^{4} (8 x) : sec^{4} (8 x) tan^{2} (8 x)

= \int sec^{4} (8 x) tan^{2} (8 x) d x

u置換積分法を適用する

= \int sec^{4} (u) tan^{2} (u) \frac{1}{8} d u

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{8} \cdot \int sec^{4} (u) tan^{2} (u) d u

三角関数の公式を使用して書き換える

= \frac{1}{8} \cdot \int (1 + tan^{2} (u)) sec^{2} (u) tan^{2} (u) d u

u置換積分法を適用する

= \frac{1}{8} \cdot \int v^{2} (1 + v^{2}) d v

拡張

v^{2} (1 + v^{2}) : v^{2} + v^{4}

= \frac{1}{8} \cdot \int v^{2} + v^{4} d v

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{8} (\int v^{2} d v + \int v^{4} d v)

\int v^{2} d v = \frac{v ^{3}}{3}

\int v^{4} d v = \frac{v ^{5}}{5}

= \frac{1}{8} (\frac{v ^{3}}{3} + \frac{v ^{5}}{5})

代用を戻す

= \frac{1}{8} (\frac{tan ^{3} ( 8 x )}{3} + \frac{tan ^{5} ( 8 x )}{5})

定数を解答に追加する

= \frac{1}{8} (\frac{tan ^{3} ( 8 x )}{3} + \frac{tan ^{5} ( 8 x )}{5}) + C

グラフ

人気の例

(\partial)/(\partial x)(2x^2ycos(x))

\frac{\partial}{\partial x} (2 x^{2} y cos (x))

d/(d{x)}(sqrt(1-{x)^2-{y}^2-{z}^2})

\frac{d}{d x} (1 - x^{2} - y^{2} - z^{2})

integral of (3x+3)sqrt(x^2+2x+3)

\int (3 x + 3) x^{2} + 2 x + 3 d x

sum from n=1 to infinity of 10(-2/3)^n

n = 1 \sum \infty 10 (- \frac{2}{3})^{n}

limit as n approaches infinity of 3/n

n \to \infty lim (\frac{3}{n})