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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^2cos(x-2x)sin(x))/(x^4)

\int \frac{x ^{2} cos ( x - 2 x ) sin ( x )}{x ^{4}} d x

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

Solution steps

\int \frac{x ^{2} cos ( x - 2 x ) sin ( x )}{x ^{4}} d x

Use the following identity:

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

= \int \frac{x ^{2} \frac{s i n ( x + x - 2 x ) + s i n ( x - ( x - 2 x ) )}{2}}{x ^{4}} 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{x ^{2} ( sin ( x + x - 2 x ) + sin ( x - ( x - 2 x ) ) )}{x ^{4}} d x

Simplify

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

= \frac{1}{2} \cdot \int \frac{sin ( 2 x )}{x ^{2}} d x

Apply Integration By Parts

: - \frac{sin ( 2 x )}{x} - \int - \frac{2 cos ( 2 x )}{x} d x

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

\int - \frac{2 cos ( 2 x )}{x} d x = - 2 Ci (2 x)

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{2 x ^{5} + 3 x ^{2}}{2 x ^{3} + 1} d x

\int \frac{( 2 )}{3 + 2 x} d x

\int (\frac{e ^{\frac{1}{x ^{3}}}}{x ^{4}}) d x

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

\int e^{6 x} x^{2} d x

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