What is the integral of cos(x)sqrt(4-sin^2(x)) ?

The integral of cos(x)sqrt(4-sin^2(x)) is 2arcsin(1/2 sin(x))+sin(2arcsin(1/2 sin(x)))+C

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integral of cos(x)sqrt(4-sin^2(x))

\int cos (x) 4 - sin^{2} (x) d x

2 arcsin (\frac{1}{2} sin (x)) + sin (2 arcsin (\frac{1}{2} sin (x))) + C

Solution steps

\int cos (x) 4 - sin^{2} (x) d x

Apply u-substitution

: \int 4 - u^{2} d u

= \int 4 - u^{2} d u

Apply Trigonometric Substitution

: \int 4 cos^{2} (v) d v

= \int 4 cos^{2} (v) d v

Take the constant out:

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

= 4 \cdot \int cos^{2} (v) d v

Rewrite using trig identities

= 4 \cdot \int \frac{1 + cos ( 2 v )}{2} d v

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \int 1 + cos (2 v) d v

Apply the Sum Rule:

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

= 4 \cdot \frac{1}{2} (\int 1 d v + \int cos (2 v) d v)

\int 1 d v = v

\int cos (2 v) d v = \frac{1}{2} sin (2 v)

= 4 \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

= 4 \cdot \frac{1}{2} (arcsin (\frac{1}{2} sin (x)) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} sin (x))))

Simplify

4 \cdot \frac{1}{2} (arcsin (\frac{1}{2} sin (x)) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} sin (x)))) : 2 arcsin (\frac{1}{2} sin (x)) + sin (2 arcsin (\frac{1}{2} sin (x)))

= 2 arcsin (\frac{1}{2} sin (x)) + sin (2 arcsin (\frac{1}{2} sin (x)))

Add a constant to the solution

= 2 arcsin (\frac{1}{2} sin (x)) + sin (2 arcsin (\frac{1}{2} sin (x))) + C

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

\int x^{9} d x

(1 - y) \cdot e^{y} \cdot y^{^{'}} + \frac{y ^{2}}{x \cdot ln ( x )} = 0

\int_{0}^{\frac{1}{14}} x ln (14 x) d x

390 - 2 p - 0.1 p^{2}

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