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As stated in the title, I am currently working on solving $$\int \cos(5x)\sinh(6x)\,dx.$$

After using integration by parts twice, I have gotten to the point where I have $$\frac{\sinh(6x)\sin(5x)}{5}+\frac{6\cosh(6x)\cos(5x)}{25}+\int \frac{-36\cos(5x)\sinh(6x)}{25}.$$

I've rewritten the first part by multiplying the first term by $5$, giving $$\frac{5\sinh(6x)\sin(5x)+6\cosh(6x)\cos(5x)}{25}+\int \frac{-36\cos(5x)\sinh(6x)}{25}.$$

I can see that the new integral is a multiple of the original, and using an online calculator I am told the solution is $$-\frac{-5\sinh(6x)\sin(5x)-6\cosh(6x)\cos(5x)}{61}+C.$$

However I can't grasp how this answer was calculated from where I am currently, though I think the $61$ in the denominator is throwing me off.

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    $\begingroup$ As $\dfrac{d(uv)}{dx}=uv'+u'v,$ and as $5\int\cos5x\ dx=\sin5x,6\int\sinh(6x)\ dx=\cosh(6x)$ Find $$\dfrac{d(A\cdot\sin5x\sinh(6x)+B\cdot\cos5x\cosh(6x))}{dx}=?$$ $\endgroup$ – lab bhattacharjee May 29 '16 at 10:20
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Let $I$ denote your original integral. You've worked out that $$ I = \frac{\sinh(6x)\sin(5x)}{5}+\frac{6\cosh(6x)\cos(5x)}{25}+\int \frac{-36\cos(5x)\sinh(6x)}{25} \\ = \frac{\sinh(6x)\sin(5x)}{5}+\frac{6\cosh(6x)\cos(5x)}{25}+ \frac{-36}{25}I. $$ From this, you know that $$ (1 + 36/25) I = \frac{\sinh(6x)\sin(5x)}{5}+\frac{6\cosh(6x)\cos(5x)}{25} $$ and you can now solve for $I$.

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  • $\begingroup$ This is likely a silly question, but where does the 1 come from? $\endgroup$ – Dwayne H May 29 '16 at 10:12
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    $\begingroup$ @DwayneH $$I + 36/25 I = (1 + 36/25) I$$ $\endgroup$ – mattos May 29 '16 at 10:27
  • $\begingroup$ Oh of course, I forgot about the $I$ that was already there. $\endgroup$ – Dwayne H May 29 '16 at 10:28

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