What is the integral of 9sin(x)cos(x) ?

The integral of 9sin(x)cos(x) is 9/2 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 9sin(x)cos(x)

\int 9 sin (x) cos (x) d x

\frac{9}{2} sin^{2} (x) + C

Solution steps

\int 9 sin (x) cos (x) d x

Take the constant out:

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

= 9 \cdot \int sin (x) cos (x) d x

Apply u-substitution

: \int u d u

= 9 \cdot \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= 9 \cdot \frac{u ^{2}}{2}

Substitute back

u = sin (x)

= 9 \cdot \frac{sin ^{2} ( x )}{2}

Simplify

9 \cdot \frac{sin ^{2} ( x )}{2} : \frac{9}{2} sin^{2} (x)

= \frac{9}{2} sin^{2} (x)

Add a constant to the solution

= \frac{9}{2} sin^{2} (x) + C

(2 e^{y} - x) y^{^{'}} = 1

\int \frac{x ^{2}}{( 225 + x ^{2} ) ^{2}} d x

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

\int \frac{4}{4 x ^{2} - 4 x + 5} d x

x \to 0 lim (2 x + x)

What is the integral of 9sin(x)cos(x) ?
The integral of 9sin(x)cos(x) is 9/2 sin^2(x)+C