What is the integral of (sin(2x))/(49+cos^2(x)) ?

The integral of (sin(2x))/(49+cos^2(x)) is -ln|cos(2x)+99|+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))/(49+cos^2(x))

\int \frac{sin ( 2 x )}{49 + cos ^{2} ( x )} d x

- ln ∣ cos (2 x) + 99 ∣ + C

Solution steps

\int \frac{sin ( 2 x )}{49 + cos ^{2} ( x )} d x

Rewrite using trig identities

= \int \frac{2 sin ( 2 x )}{cos ( 2 x ) + 99} d x

Take the constant out:

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

= 2 \cdot \int \frac{sin ( 2 x )}{cos ( 2 x ) + 99} d x

Apply u-substitution

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

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

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}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 (- \frac{1}{2} ln ∣ u ∣)

Substitute back

u = cos (2 x) + 99

= 2 (- \frac{1}{2} ln ∣ cos (2 x) + 99 ∣)

Simplify

2 (- \frac{1}{2} ln ∣ cos (2 x) + 99 ∣) : - ln ∣ cos (2 x) + 99 ∣

= - ln ∣ cos (2 x) + 99 ∣

Add a constant to the solution

= - ln ∣ cos (2 x) + 99 ∣ + C

\frac{d}{d t} (t^{3} + 3 t)

\frac{\partial}{\partial x} (y tan (x))

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

\frac{\partial}{\partial x} (2 x^{4} y + 4)

What is the integral of (sin(2x))/(49+cos^2(x)) ?
The integral of (sin(2x))/(49+cos^2(x)) is -ln|cos(2x)+99|+C