What is the integral of (cos(2t))/((1+sin(2t))^4) ?

The integral of (cos(2t))/((1+sin(2t))^4) is -1/(6(1+sin(2t))^3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (cos(2t))/((1+sin(2t))^4)

\int \frac{cos ( 2 t )}{( 1 + sin ( 2 t ) ) ^{4}} d t

- \frac{1}{6 ( 1 + sin ( 2 t ) ) ^{3}} + C

Solution steps

\int \frac{cos ( 2 t )}{( 1 + sin ( 2 t ) ) ^{4}} d t

Apply u-substitution

: \int \frac{1}{2 u ^{4}} d u

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

Apply the Power Rule

: - \frac{1}{3 u ^{3}}

= \frac{1}{2} (- \frac{1}{3 u ^{3}})

Substitute back

u = 1 + sin (2 t)

= \frac{1}{2} (- \frac{1}{3 ( 1 + sin ( 2 t ) ) ^{3}})

Simplify

\frac{1}{2} (- \frac{1}{3 ( 1 + sin ( 2 t ) ) ^{3}}) : - \frac{1}{6 ( 1 + sin ( 2 t ) ) ^{3}}

= - \frac{1}{6 ( 1 + sin ( 2 t ) ) ^{3}}

Add a constant to the solution

= - \frac{1}{6 ( 1 + sin ( 2 t ) ) ^{3}} + C

x \to - 1 lim (\frac{x + 2}{x ^{3} + 1})

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

y^{3} - (6 x + 6) + 3 x y^{2} y^{^{'}} = 0

\frac{d y}{d x} = \frac{x ^{2}}{y ^{2}}

\int 7 x^{5} d x

What is the integral of (cos(2t))/((1+sin(2t))^4) ?
The integral of (cos(2t))/((1+sin(2t))^4) is -1/(6(1+sin(2t))^3)+C