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

The integral of (1+sin^2(x))sin(2x) is sin^2(x)+1/2 sin^4(x)+C

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

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

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

Solution steps

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

Rewrite using trig identities

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

Take the constant out:

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

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

Apply u-substitution

: \int u (1 + u^{2}) d u

= 2 \cdot \int u (1 + u^{2}) d u

Expand

u (1 + u^{2}) : u + u^{3}

= 2 \cdot \int u + u^{3} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int u d u + \int u^{3} d u)

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

\int u^{3} d u = \frac{u ^{4}}{4}

= 2 (\frac{u ^{2}}{2} + \frac{u ^{4}}{4})

Substitute back

u = sin (x)

= 2 (\frac{sin ^{2} ( x )}{2} + \frac{sin ^{4} ( x )}{4})

Simplify

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

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

Add a constant to the solution

= sin^{2} (x) + \frac{1}{2} sin^{4} (x) + C

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