What is the integral of (sin(2x))/(sqrt(3-cos^4(x))) ?

The integral of (sin(2x))/(sqrt(3-cos^4(x))) is -arcsin(1/(sqrt(3))cos^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 (sin(2x))/(sqrt(3-cos^4(x)))

\int \frac{sin ( 2 x )}{3 - cos ^{4} ( x )} d x

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

Solution steps

\int \frac{sin ( 2 x )}{3 - cos ^{4} ( x )} d x

Rewrite using trig identities

= \int \frac{2 cos ( x ) sin ( x )}{3 - cos ^{4} ( x )} d x

Take the constant out:

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

= 2 \cdot \int \frac{cos ( x ) sin ( x )}{3 - cos ^{4} ( x )} d x

Apply u-substitution

: \int - \frac{u}{3 - u ^{4}} d u

= 2 \cdot \int - \frac{u}{3 - u ^{4}} d u

Take the constant out:

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

= 2 (- \int \frac{u}{3 - u ^{4}} d u)

Apply u-substitution

: \int \frac{1}{2 3 - v ^{2}} d v

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

Take the constant out:

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

= 2 (- \frac{1}{2} \cdot \int \frac{1}{3 - v ^{2}} d v)

Apply Trigonometric Substitution

: \int 1 d w

= 2 (- \frac{1}{2} \cdot \int 1 d w)

Integral of a constant:

\int a d x = a x

= 2 (- \frac{1}{2} \cdot 1 \cdot w)

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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

\int \frac{x}{2 x ^{2} + 8} d x

\int x + x^{3} d x

\int e^{2 x} e^{x} + 2 d x

\int \frac{x}{1 - 5 x ^{2}} d x

What is the integral of (sin(2x))/(sqrt(3-cos^4(x))) ?
The integral of (sin(2x))/(sqrt(3-cos^4(x))) is -arcsin(1/(sqrt(3))cos^2(x))+C