integral from-5 to cos(x) of 1/(t^2-1)

解答

\int_{- 5}^{c o s (x)} \frac{1}{t ^{2} - 1} d t

- \frac{ln ∣ cos ( x ) + 1 ∣ - ln ( \frac{2}{3} ∣ cos ( x ) - 1 ∣ )}{2}

求解步骤

\int_{- 5}^{c o s (x)} \frac{1}{t ^{2} - 1} d t

因式分解

t^{2} - 1 : - (- t^{2} + 1)

= \int_{- 5}^{c o s (x)} \frac{1}{- ( - t ^{2} + 1 )} d t

提出常数:

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

= - \int_{- 5}^{c o s (x)} \frac{1}{- t ^{2} + 1} d t

使用积分常见定则:

\int \frac{1}{- t ^{2} + 1} d t = \frac{ln ∣ t + 1 ∣}{2} - \frac{ln ∣ t - 1 ∣}{2}

= - [\frac{ln ∣ t + 1 ∣}{2} - \frac{ln ∣ t - 1 ∣}{2}]_{- 5}^{c o s (x)}

化简

- [\frac{ln ∣ t + 1 ∣}{2} - \frac{ln ∣ t - 1 ∣}{2}]_{- 5}^{c o s (x)} : - [\frac{1}{2} (ln ∣ t + 1 ∣ - ln ∣ t - 1 ∣)]_{- 5}^{c o s (x)}

= - [\frac{1}{2} (ln ∣ t + 1 ∣ - ln ∣ t - 1 ∣)]_{- 5}^{c o s (x)}

带入上下限计算后得:

\frac{1}{2} (ln ∣ cos (x) + 1 ∣ - ln ∣ cos (x) - 1 ∣) - \frac{2 ln ( 2 ) - ln ( 6 )}{2}

= - (\frac{1}{2} (ln ∣ cos (x) + 1 ∣ - ln ∣ cos (x) - 1 ∣) - \frac{2 ln ( 2 ) - ln ( 6 )}{2})

化简

= - \frac{ln ∣ cos ( x ) + 1 ∣ - ln ( \frac{2}{3} ∣ cos ( x ) - 1 ∣ )}{2}

\int_{- 2}^{2} (∣ x ∣ + x) e^{∣ x ∣} d x

\int_{1}^{4} x^{- 3} d x

\int_{7}^{8} (\frac{3}{x ^{2}} - 1) d x

\int_{0}^{5} (2 x + 5) d x

\int_{2}^{4} e^{- s t} (t^{2}) d t