What is the integral of 2^{sin^2(1/3 x)}sin(2/3 x) ?

The integral of 2^{sin^2(1/3 x)}sin(2/3 x) is (3*2^{sin^2(x/3)})/(ln(2))+C

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integral of 2^{sin^2(1/3 x)}sin(2/3 x)

\int 2^{s i n^{2} (\frac{1}{3} x)} sin (\frac{2}{3} x) d x

\frac{3 \cdot 2 ^{s i n^{2} (\frac{x}{3})}}{ln ( 2 )} + C

Solution steps

\int 2^{s i n^{2} (\frac{1}{3} x)} sin (\frac{2}{3} x) d x

Apply u-substitution

: \int 3 \cdot 2^{s i n^{2} (u)} sin (2 u) d u

= \int 3 \cdot 2^{s i n^{2} (u)} sin (2 u) d u

Take the constant out:

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

= 3 \cdot \int 2^{s i n^{2} (u)} sin (2 u) d u

Apply u-substitution

: \int 2^{v} d v

= 3 \cdot \int 2^{v} d v

\int a^{x} d x = \frac{a ^{x}}{ln a}

= 3 \cdot \frac{2 ^{v}}{ln ( 2 )}

Substitute back

= 3 \cdot \frac{2 ^{s i n^{2} (\frac{1}{3} x)}}{ln ( 2 )}

Simplify

3 \cdot \frac{2 ^{s i n^{2} (\frac{1}{3} x)}}{ln ( 2 )} : \frac{3 \cdot 2 ^{s i n^{2} (\frac{x}{3})}}{ln ( 2 )}

= \frac{3 \cdot 2 ^{s i n^{2} (\frac{x}{3})}}{ln ( 2 )}

Add a constant to the solution

= \frac{3 \cdot 2 ^{s i n^{2} (\frac{x}{3})}}{ln ( 2 )} + C

\frac{1}{x ^{2}}, - 1

\int (\frac{6 - 2 x}{8 - 4 x - 4 x ^{2}}) d x

\frac{d}{d x} (\frac{7}{x ^{4} + 3})

\frac{\partial}{\partial x} ((x - 1)^{2} + 8 (y - 1)^{2})

x \to \frac{π}{4} lim ((tan (x))^{t a n (2 x)})

What is the integral of 2^{sin^2(1/3 x)}sin(2/3 x) ?
The integral of 2^{sin^2(1/3 x)}sin(2/3 x) is (3*2^{sin^2(x/3)})/(ln(2))+C