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

The integral of sin(x)cos(x)sqrt(2-cos^2(x)) is 1/(6sqrt(2))(3-cos(2x))^{3/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sin(x)cos(x)sqrt(2-cos^2(x))

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

\frac{1}{6 2} (3 - cos (2 x))^{\frac{3}{2}} + C

Solution steps

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

Rewrite using trig identities

= \int \frac{1}{2 2} sin (2 x) 3 - cos (2 x) d x

Take the constant out:

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

= \frac{1}{2 2} \cdot \int sin (2 x) 3 - cos (2 x) d x

Apply u-substitution

: \int \frac{u}{2} d u

= \frac{1}{2 2} \cdot \int \frac{u}{2} d u

Take the constant out:

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

= \frac{1}{2 2} \cdot \frac{1}{2} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= \frac{1}{2 2} \cdot \frac{1}{2} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 3 - cos (2 x)

= \frac{1}{2 2} \cdot \frac{1}{2} \cdot \frac{2}{3} (3 - cos (2 x))^{\frac{3}{2}}

Simplify

\frac{1}{2 2} \cdot \frac{1}{2} \cdot \frac{2}{3} (3 - cos (2 x))^{\frac{3}{2}} : \frac{1}{6 2} (3 - cos (2 x))^{\frac{3}{2}}

= \frac{1}{6 2} (3 - cos (2 x))^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{6 2} (3 - cos (2 x))^{\frac{3}{2}} + C

\int (e^{s} - e^{- s})^{2} d s

\frac{5 - x ^{2}}{( 5 + x ^{2} ) ^{2}}

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

\int \frac{36}{x ^{3}} d x

x \to 1 lim (9)

What is the integral of sin(x)cos(x)sqrt(2-cos^2(x)) ?
The integral of sin(x)cos(x)sqrt(2-cos^2(x)) is 1/(6sqrt(2))(3-cos(2x))^{3/2}+C