What is the integral of (sin(θ))/(sqrt(4+cos^2(θ))) ?

The integral of (sin(θ))/(sqrt(4+cos^2(θ))) is -ln(1/2 |cos(θ)+sqrt(4+cos^2(θ))|)+C

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integral of (sin(θ))/(sqrt(4+cos^2(θ)))

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

- ln (\frac{1}{2} cos (θ) + 4 + cos^{2} (θ)) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int sec (v) d v

= - \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} cos (θ))) + sec (arctan (\frac{1}{2} cos (θ)))

Simplify

- ln tan (arctan (\frac{1}{2} cos (θ))) + sec (arctan (\frac{1}{2} cos (θ))) : - ln (\frac{1}{2} cos (θ) + 4 + cos^{2} (θ))

= - ln (\frac{1}{2} cos (θ) + 4 + cos^{2} (θ))

Add a constant to the solution

= - ln (\frac{1}{2} cos (θ) + 4 + cos^{2} (θ)) + C

\int \frac{2 x ^{2} + x + 9}{( x + 1 ) ( x ^{2} + 4 )} d x

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

\int (sin (3 x))^{7} d x

\int 3 x^{- 6} d x

What is the integral of (sin(θ))/(sqrt(4+cos^2(θ))) ?
The integral of (sin(θ))/(sqrt(4+cos^2(θ))) is -ln(1/2 |cos(θ)+sqrt(4+cos^2(θ))|)+C