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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

u = 1 + cos^{2} (x)

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

Simplify

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

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

Add a constant to the solution

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

x \to 0 lim (\frac{x ^{2} - 6 x}{x})

\frac{\partial}{\partial x} (\frac{4 x y ^{2}}{x ^{2} + y ^{4}})

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

x \to 2 lim (\frac{2 x + 1}{x - 2})

\frac{d}{d x} (sin (3 x))

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