What is the integral of 1/(4-4sin(x)) ?

The integral of 1/(4-4sin(x)) is -1/(2(tan(x/2)-1))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(4-4sin(x))

\int \frac{1}{4 - 4 sin ( x )} d x

- \frac{1}{2 ( tan ( \frac{x}{2} ) - 1 )} + C

Solution steps

\int \frac{1}{4 - 4 sin ( x )} d x

\frac{1}{4 - 4 sin ( x )} = \frac{1}{4} \frac{1}{1 - sin ( x )}

= \int \frac{1}{4} \cdot \frac{1}{1 - sin ( x )} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{1 - sin ( x )} d x

Apply u-substitution

: \int \frac{2}{1 + u ^{2} - 2 u} d u

= \frac{1}{4} \cdot \int \frac{2}{1 + u ^{2} - 2 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot 2 \cdot \int \frac{1}{1 + u ^{2} - 2 u} d u

Complete the square

1 + u^{2} - 2 u : (u - 1)^{2}

= \frac{1}{4} \cdot 2 \cdot \int \frac{1}{( u - 1 ) ^{2}} d u

Apply u-substitution

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

= \frac{1}{4} \cdot 2 \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= \frac{1}{4} \cdot 2 (- \frac{1}{v})

Substitute back

= \frac{1}{4} \cdot 2 (- \frac{1}{tan ( \frac{x}{2} ) - 1})

Simplify

\frac{1}{4} \cdot 2 (- \frac{1}{tan ( \frac{x}{2} ) - 1}) : - \frac{1}{2 ( tan ( \frac{x}{2} ) - 1 )}

= - \frac{1}{2 ( tan ( \frac{x}{2} ) - 1 )}

Add a constant to the solution

= - \frac{1}{2 ( tan ( \frac{x}{2} ) - 1 )} + C

\int 8 x^{7} e^{7 x^{8}} d x

\int \frac{- 6 x ^{3} + 8 x ^{2} - 30 x + 10}{x ^{3} + 5 x} d x

\int x^{2} + 6 d x

x ln (x)

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

What is the integral of 1/(4-4sin(x)) ?
The integral of 1/(4-4sin(x)) is -1/(2(tan(x/2)-1))+C