What is the integral of sqrt(1+1/4 sin^2(x)) ?

The integral of sqrt(1+1/4 sin^2(x)) is \E(x\vert-1)+C

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integral of sqrt(1+1/4 sin^2(x))

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

E (x ∣ - 1) + C

Solution steps

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{4} = 4^{- 1}

= \int 1 + 4^{- 1} sin^{2} (x) d x

Use the nonelementary integral:

\int 1 + c \cdot sin^{2} (x) d x = E (x ∣ - c)

= E (x ∣ - 1)

Add a constant to the solution

= E (x ∣ - 1) + C

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

x \to 6 lim (x^{2} - 3 x)

\int (\frac{1}{x - 1}) d x

\frac{\partial}{\partial x} (e^{y} \cdot sin (x))

y^{^{''}} + 12 x = 0, y (0) = - 4, y^{^{'}} (0) = 5

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