What is the integral from-5 to cos(x) of 1/(t^2-1) ?

The integral from-5 to cos(x) of 1/(t^2-1) is -(ln|cos(x)+1|-ln(2/3 |cos(x)-1|))/2

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

\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}

Solution steps

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

Factor

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

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

Take the constant out:

\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

Use the common integral:

\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)}

Simplify

- [\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)}

Compute the boundaries:

\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})

Simplify

= - \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_{1}^{9} \frac{1}{x ^{2}} d x

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

What is the integral from-5 to cos(x) of 1/(t^2-1) ?
The integral from-5 to cos(x) of 1/(t^2-1) is -(ln|cos(x)+1|-ln(2/3 |cos(x)-1|))/2