Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
2  
Here's a start: $A \sin x + B \cos x = C \sin(x+y)$, where $C^2 = A^2 + B^2$ and $\tan y = B/A$. –  Blue Sep 8 '12 at 10:00
add comment

3 Answers 3

up vote 3 down vote accepted

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.

share|improve this answer
add comment

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.

share|improve this answer
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.