What is the integral of sin^2(b)xcos(bx) ?

The integral of sin^2(b)xcos(bx) is 1/(b^2)sin^2(b)(bxsin(bx)+cos(bx))+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(b)xcos(bx)

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

\frac{1}{b ^{2}} sin^{2} (b) (b x sin (b x) + cos (b x)) + C

Solution steps

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

Take the constant out:

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

= sin^{2} (b) \cdot \int x cos (b x) d x

Apply u-substitution

: \int \frac{u cos ( u )}{b ^{2}} d u

= sin^{2} (b) \cdot \int \frac{u cos ( u )}{b ^{2}} d u

Take the constant out:

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

= sin^{2} (b) \frac{1}{b ^{2}} \cdot \int u cos (u) d u

Apply Integration By Parts

: u sin (u) - \int sin (u) d u

= sin^{2} (b) \frac{1}{b ^{2}} (u sin (u) - \int sin (u) d u)

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

= sin^{2} (b) \frac{1}{b ^{2}} (u sin (u) - (- cos (u)))

Substitute back

u = b x

= sin^{2} (b) \frac{1}{b ^{2}} (b x sin (b x) - (- cos (b x)))

Simplify

= \frac{1}{b ^{2}} sin^{2} (b) (b x sin (b x) + cos (b x))

Add a constant to the solution

= \frac{1}{b ^{2}} sin^{2} (b) (b x sin (b x) + cos (b x)) + C

\int x 2 x - 5 d x

\int \frac{3}{( 1 - 2 x ) ^{3}} d x

\int cos^{3} (x) e^{s i n (x)} d x

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

\int (ln (x) + cos (x)) d x

What is the integral of sin^2(b)xcos(bx) ?
The integral of sin^2(b)xcos(bx) is 1/(b^2)sin^2(b)(bxsin(bx)+cos(bx))+C