What is the integral from 0 to 1 of tcosh(2t) ?

The integral from 0 to 1 of tcosh(2t) is (e^4-3)/(8e^2)+1/4

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integral from 0 to 1 of tcosh(2t)

\int_{0}^{1} t cosh (2 t) d t

\frac{e ^{4} - 3}{8 e ^{2}} + \frac{1}{4}

Decimal

1.12288 \dots

Solution steps

\int_{0}^{1} t cosh (2 t) d t

Apply Integration By Parts

: [\frac{1}{2} t sinh (2 t) - \int \frac{1}{2} sinh (2 t) d t]_{0}^{1}

= [\frac{1}{2} t sinh (2 t) - \int \frac{1}{2} sinh (2 t) d t]_{0}^{1}

\int \frac{1}{2} sinh (2 t) d t = \frac{1}{4} cosh (2 t)

= [\frac{1}{2} t sinh (2 t) - \frac{1}{4} cosh (2 t)]_{0}^{1}

Compute the boundaries:

\frac{e ^{4} - 3}{8 e ^{2}} + \frac{1}{4}

= \frac{e ^{4} - 3}{8 e ^{2}} + \frac{1}{4}

\int_{0}^{1} 3 x + 4 d x

\int_{0}^{3} ∣ 4 - 2 x ∣ d x

\int_{0}^{1} x^{2} (ln (x))^{3} d x

\int_{- 4}^{1} (- x + 4) - (x^{2} + 2 x) d x

\int_{0}^{2} x^{2} - 9 x + 14 d x

What is the integral from 0 to 1 of tcosh(2t) ?
The integral from 0 to 1 of tcosh(2t) is (e^4-3)/(8e^2)+1/4