What is the integral from 2 to 4 of 1/(tsqrt(t^2-4)) ?

The integral from 2 to 4 of 1/(tsqrt(t^2-4)) is pi/6

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integral from 2 to 4 of 1/(tsqrt(t^2-4))

\int_{2}^{4} \frac{1}{t t ^{2} - 4} d t

\frac{π}{6}

Decimal

0.52359 \dots

Solution steps

\int_{2}^{4} \frac{1}{t t ^{2} - 4} d t

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{3}} \frac{1}{2} d u

= \int_{0}^{\frac{π}{3}} \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{\frac{π}{3}} 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{2} [1 \cdot u]_{0}^{\frac{π}{3}}

Simplify

= \frac{1}{2} [u]_{0}^{\frac{π}{3}}

Compute the boundaries:

\frac{π}{3}

= \frac{1}{2} \cdot \frac{π}{3}

Simplify

= \frac{π}{6}

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What is the integral from 2 to 4 of 1/(tsqrt(t^2-4)) ?
The integral from 2 to 4 of 1/(tsqrt(t^2-4)) is pi/6