What is the integral of sqrt(1+9sin^2(3x)) ?

The integral of sqrt(1+9sin^2(3x)) is 1/3 \E(3x\vert-9)+C

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integral of sqrt(1+9sin^2(3x))

\int 1 + 9 sin^{2} (3 x) d x

\frac{1}{3} E (3 x ∣ - 9) + C

Solution steps

\int 1 + 9 sin^{2} (3 x) d x

Apply u-substitution

: \int 1 + 9 sin^{2} (u) \frac{1}{3} d u

= \int 1 + 9 sin^{2} (u) \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int 1 + 9 sin^{2} (u) d u

Use the nonelementary integral:

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

= \frac{1}{3} E (u ∣ - 9)

Substitute back

u = 3 x

= \frac{1}{3} E (3 x ∣ - 9)

Add a constant to the solution

= \frac{1}{3} E (3 x ∣ - 9) + C

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

\int - e^{x} (x + 1) d x

y = \frac{x ^{2} - 4}{x}

\frac{1}{s ^{2} + s ^{2}}

x \to 6 lim (cos (x \cdot π))

What is the integral of sqrt(1+9sin^2(3x)) ?
The integral of sqrt(1+9sin^2(3x)) is 1/3 \E(3x\vert-9)+C