Integral of sin4(x)\sin^{4}{\left(x \right)}

The calculator will find the integral/antiderivative of sin4(x)\sin^{4}{\left(x \right)}, with steps shown.

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Find sin4(x)dx\int \sin^{4}{\left(x \right)}\, dx.

Solution

Rewrite the sine using the power reducing formula sin4(α)=cos(2α)2+cos(4α)8+38\sin^{4}{\left(\alpha \right)} = - \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{\cos{\left(4 \alpha \right)}}{8} + \frac{3}{8} with α=x\alpha=x:

sin4(x)dx=(cos(2x)2+cos(4x)8+38)dx{\color{red}{\int{\sin^{4}{\left(x \right)} d x}}} = {\color{red}{\int{\left(- \frac{\cos{\left(2 x \right)}}{2} + \frac{\cos{\left(4 x \right)}}{8} + \frac{3}{8}\right)d x}}}

Apply the constant multiple rule cf(x)dx=cf(x)dx\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx with c=18c=\frac{1}{8} and f(x)=4cos(2x)+cos(4x)+3f{\left(x \right)} = - 4 \cos{\left(2 x \right)} + \cos{\left(4 x \right)} + 3:

(cos(2x)2+cos(4x)8+38)dx=((4cos(2x)+cos(4x)+3)dx8){\color{red}{\int{\left(- \frac{\cos{\left(2 x \right)}}{2} + \frac{\cos{\left(4 x \right)}}{8} + \frac{3}{8}\right)d x}}} = {\color{red}{\left(\frac{\int{\left(- 4 \cos{\left(2 x \right)} + \cos{\left(4 x \right)} + 3\right)d x}}{8}\right)}}

Integrate term by term:

(4cos(2x)+cos(4x)+3)dx8=(3dx4cos(2x)dx+cos(4x)dx)8\frac{{\color{red}{\int{\left(- 4 \cos{\left(2 x \right)} + \cos{\left(4 x \right)} + 3\right)d x}}}}{8} = \frac{{\color{red}{\left(\int{3 d x} - \int{4 \cos{\left(2 x \right)} d x} + \int{\cos{\left(4 x \right)} d x}\right)}}}{8}

Apply the constant rule cdx=cx\int c\, dx = c x with c=3c=3:

4cos(2x)dx8+cos(4x)dx8+3dx8=4cos(2x)dx8+cos(4x)dx8+(3x)8- \frac{\int{4 \cos{\left(2 x \right)} d x}}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} + \frac{{\color{red}{\int{3 d x}}}}{8} = - \frac{\int{4 \cos{\left(2 x \right)} d x}}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} + \frac{{\color{red}{\left(3 x\right)}}}{8}

Apply the constant multiple rule cf(x)dx=cf(x)dx\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx with c=4c=4 and f(x)=cos(2x)f{\left(x \right)} = \cos{\left(2 x \right)}:

3x8+cos(4x)dx84cos(2x)dx8=3x8+cos(4x)dx8(4cos(2x)dx)8\frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\int{4 \cos{\left(2 x \right)} d x}}}}{8} = \frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\left(4 \int{\cos{\left(2 x \right)} d x}\right)}}}{8}

Let u=2xu=2 x.

Then du=(2x)dx=2dxdu=\left(2 x\right)^{\prime }dx = 2 dx (steps can be seen »), and we have that dx=du2dx = \frac{du}{2}.

The integral becomes

3x8+cos(4x)dx8cos(2x)dx2=3x8+cos(4x)dx8cos(u)2du2\frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\int{\cos{\left(2 x \right)} d x}}}}{2} = \frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{2}

Apply the constant multiple rule cf(u)du=cf(u)du\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du with c=12c=\frac{1}{2} and f(u)=cos(u)f{\left(u \right)} = \cos{\left(u \right)}:

3x8+cos(4x)dx8cos(u)2du2=3x8+cos(4x)dx8(cos(u)du2)2\frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{2} = \frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}}{2}

The integral of the cosine is cos(u)du=sin(u)\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}:

3x8+cos(4x)dx8cos(u)du4=3x8+cos(4x)dx8sin(u)4\frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{4} = \frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\sin{\left(u \right)}}}}{4}

Recall that u=2xu=2 x:

3x8+cos(4x)dx8sin(u)4=3x8+cos(4x)dx8sin((2x))4\frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{\sin{\left({\color{red}{u}} \right)}}{4} = \frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{\sin{\left({\color{red}{\left(2 x\right)}} \right)}}{4}

Let u=4xu=4 x.

Then du=(4x)dx=4dxdu=\left(4 x\right)^{\prime }dx = 4 dx (steps can be seen »), and we have that dx=du4dx = \frac{du}{4}.

The integral becomes

3x8sin(2x)4+cos(4x)dx8=3x8sin(2x)4+cos(u)4du8\frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \frac{{\color{red}{\int{\cos{\left(4 x \right)} d x}}}}{8} = \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{8}

Apply the constant multiple rule cf(u)du=cf(u)du\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du with c=14c=\frac{1}{4} and f(u)=cos(u)f{\left(u \right)} = \cos{\left(u \right)}:

3x8sin(2x)4+cos(u)4du8=3x8sin(2x)4+(cos(u)du4)8\frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{8} = \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{4}\right)}}}{8}

The integral of the cosine is cos(u)du=sin(u)\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}:

3x8sin(2x)4+cos(u)du32=3x8sin(2x)4+sin(u)32\frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{32} = \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \frac{{\color{red}{\sin{\left(u \right)}}}}{32}

Recall that u=4xu=4 x:

3x8sin(2x)4+sin(u)32=3x8sin(2x)4+sin((4x))32\frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left({\color{red}{u}} \right)}}{32} = \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left({\color{red}{\left(4 x\right)}} \right)}}{32}

Therefore,

sin4(x)dx=3x8sin(2x)4+sin(4x)32\int{\sin^{4}{\left(x \right)} d x} = \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left(4 x \right)}}{32}

Simplify:

sin4(x)dx=12x8sin(2x)+sin(4x)32\int{\sin^{4}{\left(x \right)} d x} = \frac{12 x - 8 \sin{\left(2 x \right)} + \sin{\left(4 x \right)}}{32}

Add the constant of integration:

sin4(x)dx=12x8sin(2x)+sin(4x)32+C\int{\sin^{4}{\left(x \right)} d x} = \frac{12 x - 8 \sin{\left(2 x \right)} + \sin{\left(4 x \right)}}{32}+C

Answer: sin4(x)dx=12x8sin(2x)+sin(4x)32+C\int{\sin^{4}{\left(x \right)} d x}=\frac{12 x - 8 \sin{\left(2 x \right)} + \sin{\left(4 x \right)}}{32}+C