What is the integral of sqrt(1+cos(4θ)) ?

The integral of sqrt(1+cos(4θ)) is 1/(sqrt(2))sin(2θ)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sqrt(1+cos(4θ))

\int 1 + cos (4 θ) d θ

\frac{1}{2} sin (2 θ) + C

Solution steps

\int 1 + cos (4 θ) d θ

Apply Integral Substitution

= \int \frac{1}{2} 1 + cos (2 u) d u

Take the constant out:

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

= \frac{1}{2} \cdot \int 1 + cos (2 u) d u

Rewrite using trig identities

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

cos^{2} (u) = (cos (u)),

assuming

cos (u) \geq 0

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

Use the common integral:

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

= \frac{1}{2} 2 sin (u)

Substitute back

u = 2 θ

= \frac{1}{2} 2 sin (2 θ)

Simplify

\frac{1}{2} 2 sin (2 θ) : \frac{1}{2} sin (2 θ)

= \frac{1}{2} sin (2 θ)

Add a constant to the solution

= \frac{1}{2} sin (2 θ) + C

\int \frac{6 x ^{2}}{( 1 - 4 x ^{3} ) ^{4}} d x

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

\int \frac{3 x ^{\frac{1}{3}}}{1 + x ^{\frac{2}{3}}} d x

\int (4 x^{2} + 2 + \frac{5}{x ^{2} + 1}) d x

\int \frac{( x + 5 ) ^{3}}{2 x} d x

What is the integral of sqrt(1+cos(4θ)) ?
The integral of sqrt(1+cos(4θ)) is 1/(sqrt(2))sin(2θ)+C