Indefinite Integrals of $trig^4(x)$
Index
\[\begin{aligned}
& \int \sin^4(x) \,\mathrm{d}x = \dfrac{3x}{8} - \dfrac{\sin(2x)}{4} + \dfrac{\sin(4x)}{32} + C \\
& \int \cos^4(x) \,\mathrm{d}x = \dfrac{3x}{8} + \dfrac{\sin(2x)}{4} + \dfrac{\sin(4x)}{32} + C \\
& \int \tan^4(x) \,\mathrm{d}x = \dfrac{\tan^3(x)}{3} - \tan(x) + x + C \\
& \int \cot^4(x) \,\mathrm{d}x = -\dfrac{\cot^3(x)}{3} + \cot(x) + x + C \\
& \int \sec^4(x) \,\mathrm{d}x = \dfrac{\tan^3(x)}{3} + \tan(x) + C \\
& \int \csc^4(x) \,\mathrm{d}x = -\dfrac{\cot^3(x)}{3} - \cot(x) + C
\end{aligned}\]
sin
\[\begin{aligned}
\sin^2(x) &= \dfrac{1-\cos(2x)}{2} \\
\therefore \int \sin^4(x) \,\mathrm{d}x &= \int \left(\dfrac{1-\cos(2x)}{2}\right)^2 \,\mathrm{d}x \\
&= \dfrac{1}{4} \int (1-\cos(2x))^2 \,\mathrm{d}x \\
&= \dfrac{1}{4} \int (1-2\cos(2x)+\cos^2(2x)) \,\mathrm{d}x \\
&= \dfrac{1}{4} \int (1-2\cos(2x)+\dfrac{1+\cos(4x)}{2}) \,\mathrm{d}x \\
&= \dfrac{1}{4} \int (\dfrac{3}{2} - 2\cos(2x) + \dfrac{\cos(4x)}{2}) \,\mathrm{d}x \\
&= \dfrac{1}{4} \int \dfrac{3}{2}\,\mathrm{d}x - \dfrac{1}{2}\int \cos(2x) \,\mathrm{d}x + \dfrac{1}{8} \int \cos(4x) \,\mathrm{d}x \\
&= \dfrac{3}{8}x - \dfrac{1}{4}\sin(2x) + \dfrac{1}{32}\sin(4x) + C
\end{aligned}\]
cos
\[\begin{aligned}
\cos^2(x) &= \dfrac{1 + \cos(2x)}{2} \\
\therefore \int \cos^4(x) \,\mathrm{d}x &= \int \left(\dfrac{1 + \cos(2x)}{2}\right)^2 \,\mathrm{d}x \\
&= \dfrac{1}{4} \int (1 + \cos(2x))^2 \,\mathrm{d}x \\
&= \dfrac{1}{4} \int (1 + 2\cos(2x) + \cos^2(2x)) \,\mathrm{d}x \\
&= \dfrac{1}{4} \int (1 + 2\cos(2x) + \dfrac{1+\cos(4x)}{2}) \,\mathrm{d}x \\
&= \dfrac{1}{4} \int (\dfrac{3}{2} + 2\cos(2x) + \dfrac{\cos(4x)}{2}) \,\mathrm{d}x \\
&= \dfrac{1}{4} \int \dfrac{3}{2}\,\mathrm{d}x + \dfrac{1}{2}\int \cos(2x) \,\mathrm{d}x + \dfrac{1}{8} \int \cos(4x) \,\mathrm{d}x \\
&= \dfrac{3}{8}x + \dfrac{1}{4}\sin(2x) + \dfrac{1}{32}\sin(4x) + C
\end{aligned}\]
tan
\[\begin{aligned}
\int \tan^2(x) \,\mathrm{d}x &= \tan(x) - x + C \\
\\
\therefore
\int \tan^4(x) \,\mathrm{d}x &= \int \tan^2(x)\tan^2(x) \,\mathrm{d}x \\
&= \int \tan^2(x)(\sec^2(x) - 1) \,\mathrm{d}x \\
&= \int (\tan^2(x)\sec^2(x) - \tan^2(x)) \,\mathrm{d}x \\
&= \int \tan^2(x)\sec^2(x) \,\mathrm{d}x - \int \tan^2(x) \,\mathrm{d}x \\
&= \int \tan^2(x) \,\mathrm{d}\tan(x) - \int \tan^2(x) \,\mathrm{d}x \\
&= \dfrac{\tan^3(x)}{3} - \tan(x) + x + C
\end{aligned}\]
cot
\[\begin{aligned}
\int \cot^2(x) \,\mathrm{d}x &= -\cot(x) - x + C \\
\\
\therefore
\int \cot^4(x) \,\mathrm{d}x &= \int \cot^2(x)\cot^2(x) \,\mathrm{d}x \\
&= \int \cot^2(x)(\csc^2(x) - 1) \,\mathrm{d}x \\
&= \int (\cot^2(x)\csc^2(x) - \cot^2(x)) \,\mathrm{d}x \\
&= \int \cot^2(x)\csc^2(x) \,\mathrm{d}x - \int \cot^2(x) \,\mathrm{d}x \\
&= -\int \cot^2(x) \,\mathrm{d}\cot(x) - \int \cot^2(x) \,\mathrm{d}x \\
&= -\dfrac{\cot^3(x)}{3} + \cot(x) + x + C
\end{aligned}\]
sec
\[\begin{aligned}
\dfrac{\mathrm{d}}{\mathrm{d}x} \tan(x) &= \sec^2(x) \\
\int \sec^2(x) \,\mathrm{d}x &= \tan(x) + C \\
\\
\therefore
\int \sec^4(x) \,\mathrm{d}x &= \int \sec^2(x)\sec^2(x) \,\mathrm{d}x \\
&= \int \sec^2(x)(\tan^2(x) + 1) \,\mathrm{d}x \\
&= \int (\sec^2(x)\tan^2(x) + \sec^2(x)) \,\mathrm{d}x \\
&= \int \sec^2(x)\tan^2(x) \,\mathrm{d}x + \int \sec^2(x) \,\mathrm{d}x \\
&= \int \tan^2(x) \,\mathrm{d}\tan(x) + \int \sec^2(x) \,\mathrm{d}x \\
&= \dfrac{\tan^3(x)}{3} + \tan(x) + C
\end{aligned}\]
csc
\[\begin{aligned}
\dfrac{\mathrm{d}}{\mathrm{d}x} \cot(x) &= -\csc^2(x) \\
\int \csc^2(x) \,\mathrm{d}x &= -\cot(x) + C \\
\\
\therefore
\int \csc^4(x) \,\mathrm{d}x &= \int \csc^2(x)\csc^2(x) \,\mathrm{d}x \\
&= \int \csc^2(x)(\cot^2(x) + 1) \,\mathrm{d}x \\
&= \int (\csc^2(x)\cot^2(x) + \csc^2(x)) \,\mathrm{d}x \\
&= \int \csc^2(x)\cot^2(x) \,\mathrm{d}x + \int \csc^2(x) \,\mathrm{d}x \\
&= -\int \cot^2(x) \,\mathrm{d}\cot(x) + \int \csc^2(x) \,\mathrm{d}x \\
&= -\dfrac{\cot^3(x)}{3} - \cot(x) + C
\end{aligned}\]