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

The integral of sin^2(x)cos(4x) is -1/8 (1/2 sin(2x)-1/6 sin(6x))+1/4 sin^2(x)sin(4x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

- \frac{1}{8} (\frac{1}{2} sin (2 x) - \frac{1}{6} sin (6 x)) + \frac{1}{4} sin^{2} (x) sin (4 x) + C

Solution steps

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

Rewrite using trig identities

= \int (1 - cos^{2} (x)) cos (4 x) d x

Apply Integration By Parts

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

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

\int \frac{1}{4} sin (2 x) sin (4 x) d x = \frac{1}{8} (\frac{1}{2} sin (2 x) - \frac{1}{6} sin (6 x))

= - \frac{1}{8} (\frac{1}{2} sin (2 x) - \frac{1}{6} sin (6 x)) + \frac{1}{4} sin^{2} (x) sin (4 x)

Add a constant to the solution

= - \frac{1}{8} (\frac{1}{2} sin (2 x) - \frac{1}{6} sin (6 x)) + \frac{1}{4} sin^{2} (x) sin (4 x) + C

\int (3 - 4 x)^{9} d x

\int cos (θ) e^{s i n (θ)} d θ

\int (5 sin (x)) d x

\int 2 t e^{2 t} + \frac{1}{3} e^{t} d t

\int \frac{sin ( 2 x )}{6 + cos ^{2} ( x )} d x

What is the integral of sin^2(x)cos(4x) ?
The integral of sin^2(x)cos(4x) is -1/8 (1/2 sin(2x)-1/6 sin(6x))+1/4 sin^2(x)sin(4x)+C