Use double angle an pythagorean identities to manipulate before integrating.
", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "Use the hyperbolic identities to evaluate the following integrals.
\nDon't forget to include '$+C$'!
", "advice": "a)
\nWe recall the 'double angle' identity:
\n$$
\\cosh^2(x) = \\frac{1}{2} \\left( 1 + \\cosh\\left(2x\\right) \\right).
$$
Hence,
\n$$
\\begin{aligned}
\\int \\var{A1} \\cosh(\\var{B1}x)^2 + \\var{C1} \\sinh({\\simplify{2*{B1}}}x) \\, dx &= \\int \\frac{\\var{A1}}{2}\\left(1+\\cosh(\\simplify{{B1}x/2}) \\right)+ {C1} sinh({2*B1}x) \\, dx \\\\
&= \\int \\simplify{{A1}/2 + {A1}/2 cosh ({2*B1}x) + {C1}sinh ({2B1}x)} dx \\\\
&= \\simplify[basic, simplifyfractions, collectnumbers, unitdenominator, unitfactor]{{A1}x/2 + {A1}/{4*B1}sinh({2*B1}x) + {C1}/{2*B1}cosh({2B1}x) + C}.\\\\
\\end{aligned}
$$
b
\nWe recall the identity
\n$$
\\cosh^2(x) - \\sinh^2(x) = 1.
$$
Then
\n$$
\\begin{aligned}
\\int \\sinh^5(x) - \\sinh(x) \\, dx &= \\int \\sinh^4(x) \\sinh(x) - \\sinh(x) \\, dx \\\\
&= \\int(\\cosh^2(x) - 1)^2\\sinh(x) - \\sinh(x) \\, dx \\\\
&= \\int (\\cosh^4(x) - 2\\cosh^2(x) + 1)\\sinh(x) - \\sinh(x) \\, dx \\\\
&= \\int \\cosh^4(x)\\sinh(x) - 2\\cosh^2(x)\\sinh(x) \\, dx \\\\
&= \\simplify{1/5cosh(x)^5 + 2/3cosh(x)^3 + C}.
\\end{aligned}
$$
$$
\\begin{aligned}
\\int \\sinh^3(x) - \\sinh(x) \\, dx &= \\int \\sinh^2(x) \\sinh(x) - \\sinh(x) \\, dx \\\\
&= \\int(\\cosh^2(x) - 1)\\sinh(x) - \\sinh(x) \\, dx \\\\
&= \\int \\cosh^2(x)\\sinh(x) - 2\\sinh(x) \\, dx \\\\
&= \\simplify{1/3cosh(x)^3 - 2cosh(x) + C}.
\\end{aligned}
$$
\n
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$y =\\simplify{ {A1} cosh({B1}x)^2 + {C1} sinh({2*B1}x)}$
\n$\\int y \\, dx= $ [[0]]
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\n$ \\int y \\, dx = $ [[0]]
\nNote: to input a function to a power such as $\\sinh^3(x)$, type sinh(x)^3 (similarly for the other functions).