What is the integral of t^{-2}sin^2(1+1/t) ?

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

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integral of t^{-2}sin^2(1+1/t)

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

- \frac{1}{2} (1 + \frac{1}{t} - \frac{1}{2} sin (2 (1 + \frac{1}{t}))) + C

Solution steps

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

Apply u-substitution

: \int - sin^{2} (u) d u

= \int - sin^{2} (u) d u

Take the constant out:

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

= - \int sin^{2} (u) d u

Rewrite using trig identities

= - \int \frac{1 - cos ( 2 u )}{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 1 - cos (2 u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \frac{1}{2} (\int 1 d u - \int cos (2 u) d u)

\int 1 d u = u

\int cos (2 u) d u = \frac{1}{2} sin (2 u)

= - \frac{1}{2} (u - \frac{1}{2} sin (2 u))

Substitute back

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

= - \frac{1}{2} (1 + \frac{1}{t} - \frac{1}{2} sin (2 (1 + \frac{1}{t})))

Add a constant to the solution

= - \frac{1}{2} (1 + \frac{1}{t} - \frac{1}{2} sin (2 (1 + \frac{1}{t}))) + C

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