What is the integral of 1/(t^2)cos(1/t-1) ?

The integral of 1/(t^2)cos(1/t-1) is -sin(1/t-1)+C

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integral of 1/(t^2)cos(1/t-1)

\int \frac{1}{t ^{2}} cos (\frac{1}{t} - 1) d t

- sin (\frac{1}{t} - 1) + C

Solution steps

\int \frac{1}{t ^{2}} cos (\frac{1}{t} - 1) d t

Apply u-substitution

: \int - cos (u) d u

= \int - cos (u) d u

Take the constant out:

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

= - \int cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= - sin (u)

Substitute back

u = \frac{1}{t} - 1

= - sin (\frac{1}{t} - 1)

Add a constant to the solution

= - sin (\frac{1}{t} - 1) + C

y = \frac{4}{x ^{5}}

f (x) = 2 x^{3} ln (x)

\int \frac{1}{a ^{2} + v ^{2}} d v

\frac{d}{d x} (\frac{3}{4} x^{\frac{4}{3}})

n = 1 \sum \infty \frac{6 n}{2 n + 1}

What is the integral of 1/(t^2)cos(1/t-1) ?
The integral of 1/(t^2)cos(1/t-1) is -sin(1/t-1)+C