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I want to know how to integrate this function I have tried many things many substitutions but none works.

I even try to expand the numerator by $\sin3x$ and $\cos3x$ properties and tried to convert higher powers to multiple angles

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3 Answers 3

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$$ \int \frac{\sin {3x} + \cos{3x}}{ \sin^3 x + \cos^3 x } dx\ =\int \frac{3\sin x -4\sin^3x+4\cos^3x-3\cos x}{ \sin^3 x + \cos^3 x } dx=\int \frac{(\cos x -\sin x)(1+ 4\sin x \cos x)}{ (\sin x + \cos x)(1- \sin x\cos x) } dx=\int \frac{(\cos^2 x -\sin^2 x)(1+ 4\sin x \cos x)}{ (\sin x + \cos x)^2(1- \sin x\cos x) } dx=\int \frac{\cos 2x (1+ 2\sin 2x)}{ (1+ \sin 2x)(1- 0.5\sin 2x) } dx=0.5\int \frac{1+ 2\sin 2x}{ (1+ \sin 2x)(1- 0.5\sin 2x) } d(\sin 2x)$$ Hope you can finish from here.

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  • $\begingroup$ How does $3\sin x - 4\sin^3 x+4\cos^3 x- 3\cos x$ become $(\cos x -\sin x)(1+\sin x\cos x)$? Shouldn't it be $(\cos x -\sin x)(1+4\sin x\cos x)$? $\endgroup$
    – GoodDeeds
    Sep 4, 2018 at 20:11
  • $\begingroup$ @GoodDeeds: Thank you for spotting my mistake $3(\sin x-\cos x)-4(\sin x-\cos x)(1+\sin x \cos x)=(\sin x - \cos x)(3-4(1+\sin x \cos x))=(\sin x - \cos x)(3-4-4\sin x \cos x)=(\cos x - \sin x)(1+ 4\sin x\cos x)$ $\endgroup$
    – Vasili
    Sep 4, 2018 at 20:25
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Use \begin{equation}\cos\left(3x\right)=\cos^3\left(x\right)-3\cos\left(x\right)\sin^2\left(x\right) \end{equation} and \begin{equation} \sin\left(3x\right)=3\cos^2\left(x\right)\sin\left(x\right)-\sin^3\left(x\right) \end{equation} accompanied with \begin{align} \sin\left(x\right)&=\dfrac{\tan\left(x\right)}{\sec\left(x\right)} \\ \cos\left(x\right)&=\dfrac{1}{\sec\left(x\right)} \\ \sec^2\left(x\right)&=\tan^2\left(x\right)+1 \end{align} You will get \begin{equation} {\int}{{\sec^2\left(x\right)}}{{\left(-\dfrac{\tan^3\left(x\right)+3\tan^2\left(x\right)-3\tan\left(x\right)-1}{\tan^5\left(x\right)+\tan^3\left(x\right)+\tan^2\left(x\right)+1}\right)}}\,\mathrm{d}x \end{equation} Use the change of variable $u = \tan (x)$ then $dx = \frac{1}{\sec^2 (x)} du$, you will get \begin{equation} {\int}\dfrac{u^3+3u^2-3u-1}{u^5+u^3+u^2+1}\,\mathrm{d}u \end{equation} But \begin{equation} {\int}\dfrac{u^3+3u^2-3u-1}{u^5+u^3+u^2+1}\,\mathrm{d}u ={\int}\dfrac{u^3+3u^2-3u-1}{\left(u+1\right)\left(u^2+1\right)\left(u^2-u+1\right)}\,\mathrm{d}u \end{equation} Now perform partial fraction decomposition, we get \begin{equation} {\int}\dfrac{u^3+3u^2-3u-1}{u^5+u^3+u^2+1}\,\mathrm{d}u ={\int}\left(\dfrac{10u-5}{3\left(u^2-u+1\right)}-\dfrac{4u}{u^2+1}+\dfrac{2}{3\left(u+1\right)}\right)\mathrm{d}u \end{equation} That is \begin{equation} {\int}\dfrac{u^3+3u^2-3u-1}{u^5+u^3+u^2+1}\,\mathrm{d}u =\dfrac{5}{3}\underbrace{{}{\int}\dfrac{2u-1}{u^2-u+1}\,\mathrm{d}u}_{\ln \vert u^2-u+1 \vert}- \underbrace{{\int}\dfrac{u}{u^2+1}\,\mathrm{d}u}_{\dfrac{\ln\vert u^2+1\vert}{2}}+ {{\dfrac{2}{3}}} \underbrace{{\int}\dfrac{1}{u+1}\,\mathrm{d}u}_{\ln\vert u+1\vert} \end{equation}

Finally plug back $u = \tan x$.

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$$I=\int \frac{\sin {3x} + \cos{3x}}{ \sin^3 x + \cos^3 x } dx=\cdots=\int \frac{(\cos x -\sin x)(1+ 4\sin x \cos x)}{ (\sin x + \cos x)(1- \sin x\cos x) } dx$$

As $\dfrac{d(\sin x+\cos x)}{dx}=\cos x-\sin x$

if we choose $\sin x+\cos x=u, 2\sin x\cos x=u^2-1$

$$I=2\int\dfrac{1+2(u^2-1)}{u(2-(u^2-1))}du$$

Now use Partial Fraction Decomposition, $$\dfrac{2u^2-1}{u(3-u^2)}=\dfrac Au+\dfrac{Bu+C}{3-u^2}$$

$\implies3A=-1,B-A=2$

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