What is the integral of (sin(θ))/(cos^3(θ)+cos(θ)) ?

The integral of (sin(θ))/(cos^3(θ)+cos(θ)) is -ln|cos(θ)|+1/2 ln|cos^2(θ)+1|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (sin(θ))/(cos^3(θ)+cos(θ))

\int \frac{sin ( θ )}{cos ^{3} ( θ ) + cos ( θ )} d θ

- ln ∣ cos (θ) ∣ + \frac{1}{2} ln cos^{2} (θ) + 1 + C

Solution steps

\int \frac{sin ( θ )}{cos ^{3} ( θ ) + cos ( θ )} d θ

Apply u-substitution

: \int - \frac{1}{u ^{3} + u} d u

= \int - \frac{1}{u ^{3} + u} d u

Take the constant out:

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

= - \int \frac{1}{u ^{3} + u} d u

Take the partial fraction of

\frac{1}{u ^{3} + u} : \frac{1}{u} - \frac{u}{u ^{2} + 1}

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{u}{u ^{2} + 1} d u = \frac{1}{2} ln u^{2} + 1

= - (ln ∣ u ∣ - \frac{1}{2} ln u^{2} + 1)

Substitute back

u = cos (θ)

= - (ln ∣ cos (θ) ∣ - \frac{1}{2} ln cos^{2} (θ) + 1)

Simplify

= - ln ∣ cos (θ) ∣ + \frac{1}{2} ln cos^{2} (θ) + 1

Add a constant to the solution

= - ln ∣ cos (θ) ∣ + \frac{1}{2} ln cos^{2} (θ) + 1 + C

cos (2 t)

x \to - 3 lim (- x^{2} + 5 x - 1)

x \to + 0 lim (x^{2} cos (\frac{1}{x}))

\frac{d}{d x} (- 3 e^{- 2 x})

\int_{0}^{4} \frac{6 t}{( t - 5 ) ^{2}} d t

What is the integral of (sin(θ))/(cos^3(θ)+cos(θ)) ?
The integral of (sin(θ))/(cos^3(θ)+cos(θ)) is -ln|cos(θ)|+1/2 ln|cos^2(θ)+1|+C