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I need some hints, suggestions for the following integral $$\int_{\pi/4}^{\pi/2} \frac{x^2+8}{(x^2-16)\sin (x) + 8 x \cos(x)} \ dx$$

Since it's a high school problem, I thought of some variable change, integration by parts, but I can't see yet how to make them work. I don't know where I should start from. Thanks!

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    $\begingroup$ Here's a start: $A \sin x + B \cos x = C \sin(x+y)$, where $C^2 = A^2 + B^2$ and $\tan y = B/A$. $\endgroup$ – Blue Sep 8 '12 at 10:00
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Let $u = 2\tan^{-1}\left(\frac{x}{4}\right) - x$. Then

$$ du = -\frac{x^2 + 8}{x^2 + 16}\,dx. $$

Now since

$$ \begin{align*} \frac{x^2 + 8}{(x^2 - 16)\sin x + 8x \cos x} &= \frac{\frac{x^2 + 8}{x^2 + 16}}{\frac{8x}{16+x^2} \cos x - \frac{16-x^2}{16+x^2}\sin x} \\ &= \frac{\frac{x^2 + 8}{x^2 + 16}}{\sin\left(2\tan^{-1}\left(\frac{x}{4}\right)\right) \cos x - \cos\left(2\tan^{-1}\left(\frac{x}{4}\right)\right)\sin x} \\ &= \frac{\frac{x^2 + 8}{x^2 + 16}}{\sin\left(2\tan^{-1}\left(\frac{x}{4}\right) - x\right)} \\ &= -\frac{1}{\sin u}\frac{du}{dx}, \end{align*}$$

it remains to take integration by substitution.

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Following up on Blue's comment:

In general $$ a\sin x+b\cos x = \sqrt{a^2+b^2}\Big( \frac{a}{\sqrt{a^2+b^2}}\sin x + \frac{b}{\sqrt{a^2+b^2}}\cos x \Big) = \sqrt{a^2+b^2} \left(\cos\varphi\sin x+\sin\varphi\cos x\right) = \sqrt{a^2+b^2}\sin(x+\varphi), $$ so $\tan\varphi = \dfrac b a$.

Now putting in $a=x^2-16$ and $b=8x$, we have $$ \begin{align} \Big( x^2 - 16 \Big)^2 + \Big(8x\Big)^2 & = \Big(x^4 - 32x^2 + 256\Big) + \Big(64x^2\Big) \\[10pt] & = x^4 + 32x^2 + 256 = \Big( x^2 + 16 \Big)^2 \end{align} $$ Therefore $$ (x^2-16)\sin x + 8x\cos x = (x^2+16)\sin(x+\varphi) $$ where $$ \varphi = \arctan\frac{8x}{x^2-16}. \tag{1} $$

The answer from sos440 doesn't explain how s/he thought of that susbstitution in the first place, but conjoining that with $(1)$, I'm thinking: let's see if that last fraction in $(1)$ is the tangent of a double angle. Remember that $$ \tan(2\alpha) = \frac{2\tan\alpha}{1-\tan^2\alpha}. $$ Therefore $$ \frac{8x}{x^2-16} = \frac{2(-x/4)}{1-(-x/4)^2} = \tan(2\alpha)\text{ where } \tan\alpha = \frac{-x}{4}. $$ After that, proceed as in the answer from sos440.

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Let $R\sin A=x^2-16$ and $R\cos A=8x\implies R=x^2+16$ and $\cos A=\frac{8x}{x^2+16}$

So, $(x^2-16)\sin x+8x\cos x$

$=(x^2+16)(\cos A \cos x+\sin A\sin x)$

$=(x^2+16)\cos (x-A)$

$=(x^2+16)\cos(x-\cos^{-1}(\frac{8x}{x^2+16}))$

Putting $y=x-\cos^{-1}(\frac{8x}{x^2+16})$, we get $\frac{dy}{dx}=\frac{x^2+8}{x^+16}$

$$\int\frac{x^2+8}{(x^2-16)\sin (x) + 8 x \cos(x)} \ dx$$

$$=\int\frac{x^2+8}{(x^2-16)\cos(x-\cos^{-1}(\frac{8x}{x^2+16}))} \ dx$$

$$=\int\frac{dy}{\mathrm{cos}y}=\int \sec ydy=\ln|\sec y+ \tan y|+C=\ln\tan (\frac{\pi}{4}+\frac{y}{2})+C$$ where C is the indeterminate constant for indefinite integral.

Now if $2B=\cos^{-1}(\frac{8x}{x^2+16}),\frac{8x}{x^2+16}=\cos2B$ $=\frac{1-\tan^2B}{1+\tan^2B}\implies \tan^2B=(\frac{4+x}{4-x})^2$

$y=x-2\tan^{-1}(\frac{4+x}{4-x})=x-2(\frac{\pi}{4}+\tan^{-1}(\frac{x}{4})) $

When $x=\frac{\pi}{2}, y=-2\tan^{-1}\frac{\pi}{8}$

When $x=\frac{\pi}{4}, y=-\frac{\pi}{4}-2\tan^{-1}\frac{\pi}{16}$

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