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

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

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integral of+(sin^2(1/t))/(t^2)

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

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

Solution steps

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

Rewrite using trig identities

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

Take the constant out:

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

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

Expand

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

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

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

\int \frac{1}{t ^{2}} d t = - \frac{1}{t}

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

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

Simplify

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

Add a constant to the solution

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

n = 1 \sum \infty \frac{n !}{n 3 ^{n}}

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\frac{d}{d x} (\frac{5 tan ( x )}{x})

\int \frac{7}{5 y - 105} d y

\frac{\partial}{\partial x} (x^{2} + x y - 25)

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