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

The integral of 1/(x^2)cos^2(1/x) is -1/4 (2/x+sin(2/x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(x^2)cos^2(1/x)

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

- \frac{1}{4} (\frac{2}{x} + sin (\frac{2}{x})) + C

Solution steps

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

Rewrite using trig identities

= \int \frac{1 + cos ( \frac{2}{x} )}{2 x ^{2}} d x

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}{x} )}{x ^{2}} d x

Apply u-substitution

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

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

\int 1 d u = u

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

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

Substitute back

u = \frac{2}{x}

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

Simplify

\frac{1}{2} (- \frac{1}{2} (\frac{2}{x} + sin (\frac{2}{x}))) : - \frac{1}{4} (\frac{2}{x} + sin (\frac{2}{x}))

= - \frac{1}{4} (\frac{2}{x} + sin (\frac{2}{x}))

Add a constant to the solution

= - \frac{1}{4} (\frac{2}{x} + sin (\frac{2}{x})) + C

f (x) = \frac{2 x}{x}

\int 7 tan^{2} (x se) c^{3} x

\frac{\partial}{\partial x} (\frac{x y}{x + y} + z)

\frac{d}{d x} (\frac{( 2 x + 3 ) ^{4}}{( 3 x - 2 ) ^{5}})

\int_{a}^{b} arctan (x) d x

What is the integral of 1/(x^2)cos^2(1/x) ?
The integral of 1/(x^2)cos^2(1/x) is -1/4 (2/x+sin(2/x))+C