What is the integral from-i to 3i of sin(t) ?

The integral from-i to 3i of sin(t) is -cos(3i)+cos(i)

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integral from-i to 3i of sin(t)

\int_{- i}^{3 i} sin (t) d t

- cos (3 i) + cos (i)

Solution steps

\int_{- i}^{3 i} sin (t) d t

Use the common integral:

\int sin (t) d t = - cos (t)

= [- cos (t)]_{- i}^{3 i}

Compute the boundaries:

- cos (3 i) + cos (i)

= - cos (3 i) + cos (i)

\int_{1}^{x^{2}} 16 x y^{3} d y

\int_{0}^{3} (t^{2} + 2 t - 3) d t

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

\int_{- 1}^{2} ∣ 2 x - 1 ∣ d x

\int_{- 1}^{1} \frac{x}{2 + x ^{2}} d x

What is the integral from-i to 3i of sin(t) ?
The integral from-i to 3i of sin(t) is -cos(3i)+cos(i)