What is the integral of xcos^2(2x) ?

The integral of xcos^2(2x) is 1/32 (8x^2+4xsin(4x)+cos(4x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of xcos^2(2x)

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

\frac{1}{32} (8 x^{2} + 4 x sin (4 x) + cos (4 x)) + C

Solution steps

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

Rewrite using trig identities

= \int \frac{1}{2} x (1 + cos (4 x)) 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 x (1 + cos (4 x)) d x

Apply Integration By Parts

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

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

\int x + \frac{1}{4} sin (4 x) d x = \frac{x ^{2}}{2} - \frac{1}{16} cos (4 x)

= \frac{1}{2} (x (x + \frac{1}{4} sin (4 x)) - (\frac{x ^{2}}{2} - \frac{1}{16} cos (4 x)))

Simplify

\frac{1}{2} (x (x + \frac{1}{4} sin (4 x)) - (\frac{x ^{2}}{2} - \frac{1}{16} cos (4 x))) : \frac{1}{32} (8 x^{2} + 4 x sin (4 x) + cos (4 x))

= \frac{1}{32} (8 x^{2} + 4 x sin (4 x) + cos (4 x))

Add a constant to the solution

= \frac{1}{32} (8 x^{2} + 4 x sin (4 x) + cos (4 x)) + C

\frac{( - 1 )}{s ^{2} + 5 s + 6}

f (x) = \frac{3 x - 1}{x}

\int \frac{cos ^{5} ( x )}{sin ^{3} ( x )} d x

\int_{0}^{0.6} \frac{1}{( 1 - x ) ( 3 - x )} d x

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

What is the integral of xcos^2(2x) ?
The integral of xcos^2(2x) is 1/32 (8x^2+4xsin(4x)+cos(4x))+C