What is the integral of sin(2t)tan(2t) ?

The integral of sin(2t)tan(2t) is 1/2 (ln|tan(2t)+sec(2t)|-sin(2t))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sin(2t)tan(2t)

\int sin (2 t) tan (2 t) d t

\frac{1}{2} (ln ∣ tan (2 t) + sec (2 t) ∣ - sin (2 t)) + C

Solution steps

\int sin (2 t) tan (2 t) d t

Rewrite using trig identities

= \int \frac{1 - cos ^{2} ( 2 t )}{cos ( 2 t )} d t

Apply u-substitution

: \int \frac{sin ( u ) tan ( u )}{2} d u

= \int \frac{sin ( u ) tan ( 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} \cdot \int sin (u) tan (u) d u

Rewrite using trig identities

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

Expand

\frac{1 - cos ^{2} ( u )}{cos ( u )} : \frac{1}{cos ( u )} - cos (u)

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{cos ( u )} d u = ln ∣ tan (u) + sec (u) ∣

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

= \frac{1}{2} (ln ∣ tan (u) + sec (u) ∣ - sin (u))

Substitute back

u = 2 t

= \frac{1}{2} (ln ∣ tan (2 t) + sec (2 t) ∣ - sin (2 t))

Add a constant to the solution

= \frac{1}{2} (ln ∣ tan (2 t) + sec (2 t) ∣ - sin (2 t)) + C

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

\int cos (7 x y) d x

\int (3 - 4 x)^{5} d x

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

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

What is the integral of sin(2t)tan(2t) ?
The integral of sin(2t)tan(2t) is 1/2 (ln|tan(2t)+sec(2t)|-sin(2t))+C