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

The integral of sin(2x)sqrt(1+sin^2(x)) is 1/(3sqrt(2))(3-cos(2x))^{3/2}+C

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

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

\frac{1}{3 2} (3 - cos (2 x))^{\frac{3}{2}} + C

Solution steps

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

Rewrite using trig identities

= \int 1 + sin^{2} (x) \cdot 2 cos (x) sin (x) d x

Take the constant out:

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

= 2 \cdot \int 1 + sin^{2} (x) cos (x) sin (x) d x

Rewrite using trig identities

= 2 \cdot \int \frac{1}{2 2} sin (2 x) 3 - cos (2 x) d x

Take the constant out:

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

= 2 \cdot \frac{1}{2 2} \cdot \int sin (2 x) 3 - cos (2 x) d x

Apply u-substitution

: \int \frac{u}{2} d u

= 2 \cdot \frac{1}{2 2} \cdot \int \frac{u}{2} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2 2} \cdot \frac{1}{2} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= 2 \cdot \frac{1}{2 2} \cdot \frac{1}{2} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 3 - cos (2 x)

= 2 \cdot \frac{1}{2 2} \cdot \frac{1}{2} \cdot \frac{2}{3} (3 - cos (2 x))^{\frac{3}{2}}

Simplify

2 \cdot \frac{1}{2 2} \cdot \frac{1}{2} \cdot \frac{2}{3} (3 - cos (2 x))^{\frac{3}{2}} : \frac{1}{3 2} (3 - cos (2 x))^{\frac{3}{2}}

= \frac{1}{3 2} (3 - cos (2 x))^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{3 2} (3 - cos (2 x))^{\frac{3}{2}} + C

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