What is the integral of (cos(θ))/(sqrt(2+sin^2(θ))) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (cos(θ))/(sqrt(2+sin^2(θ)))

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

ln \frac{1}{2} sin (θ) + \frac{1}{2} (2 + sin^{2} (θ)) + C

Solution steps

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

Rewrite using trig identities

= \int \frac{cos ( θ )}{- cos ^{2} ( θ ) + 3} d θ

Apply u-substitution

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

Apply Trigonometric Substitution

= \int sec (v) d v

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= ln ∣ tan (v) + sec (v) ∣

Substitute back

= ln tan (arctan (\frac{1}{2} sin (θ))) + sec (arctan (\frac{1}{2} sin (θ)))

Simplify

ln tan (arctan (\frac{1}{2} sin (θ))) + sec (arctan (\frac{1}{2} sin (θ))) : ln \frac{1}{2} sin (θ) + \frac{1}{2} (2 + sin^{2} (θ))

= ln \frac{1}{2} sin (θ) + \frac{1}{2} (2 + sin^{2} (θ))

Add a constant to the solution

= ln \frac{1}{2} sin (θ) + \frac{1}{2} (2 + sin^{2} (θ)) + C

x \to 0 lim (\frac{4 x ^{2}}{x})

\frac{\partial}{\partial x} ((41 - x^{2} - y^{2})^{\frac{1}{2}})

\int_{0}^{1} x e^{- 4 x} d x

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

d er i v a t i v e \frac{x ^{6}}{6} + \frac{1}{16 x ^{4}}

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