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Let us consider the integral
$$\int \frac{x^2}{(x\sin x+\cos x)^2}\, dx$$

I have tried with the following way
$\displaystyle \int \frac{x^2}{(x\sin x+\cos x)^2}\, dx$
$\displaystyle \Rightarrow \int \frac{x^2\sec^2 x}{(x\tan x+1)^2}\, dx$.
How can I do the next step.

juantheron
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MKSar
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    Similar question:http://math.stackexchange.com/questions/1221339/how-to-integrate-int-fracx2x-sinx-cosx2-mathrmdx –  Aug 23 '15 at 18:06

2 Answers2

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Let $$\displaystyle I = \int\frac{x^2}{(x\sin x+\cos x)^2}dx$$

We can write

$$\displaystyle (x\sin x+\cos x) = \sqrt{1+x^2}\left\{\frac{x}{\sqrt{1+x^2}}\cdot \sin x+\frac{1}{\sqrt{1+x^2}}\cdot \cos x\right\} = \sqrt{1+x^2}\cdot \cos \left(x-\phi \right)$$

So Here $$\displaystyle \cos x\ \phi = \frac{1}{\sqrt{1+x^2}}$$ and $\displaystyle \sin \phi = \frac{x}{\sqrt{1+x^2}}$ and $\tan \phi = x\Rightarrow \phi = \tan^{-1}(x)$

So Integral

$$\displaystyle I = \int\frac{x^2}{(1+x^2)\cdot \cos^2(x-\phi)}dx = \int \sec^2 (x-\phi)\cdot \frac{x^2}{1+x^2} dx = \int \sec^2 (x-\tan^{-1}(x))\cdot \frac{x^2}{(1+x^2)}dx$$

Now Let $$\left(x-\tan^{-1}x\right) = t\;,$$ Then $\displaystyle \left(1-\frac{1}{1+x^2}\right)dx = dt\Rightarrow \frac{x^2}{1+x^2}dx = dt$

So $$\displaystyle I = \int \sec^2 t dt = \tan t +\mathcal {C} = \tan \left(x-\tan^{-1} x\right)+\mathcal{C} = \left(\frac{\tan x-x}{1+x\cdot \tan x}\right)+\mathcal{C}$$

So $$\displaystyle I = \int\frac{x^2}{(x\sin x+\cos x)^2}dx = \left(\frac{\sin x- x\cos x}{\cos x+x\sin x}\right)+\mathcal{C}$$

juantheron
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1

$\bf{Another\; Solution::}$ Let $$\displaystyle I = \int \frac{x^2}{(x\sin x+\cos x)^2}dx$$

Put $x=\tan \phi\;,$ Then $dx = \sec^2 \phi d\phi$

So Integral $$\displaystyle I = \int\frac{\tan^2 \phi \cdot \sec^2 \phi}{\left[\tan \phi\cdot \sin (\tan \phi)+\cos(\tan \phi)\vphantom{\frac11}\right]^2}d\phi $$

So $$\displaystyle I = \int\frac{\tan ^2\phi \cdot \sec^2 \phi\cdot \cos^2 \phi}{\left[\sin \phi\cdot \sin (\tan \phi)+\cos \phi\cos(\tan \phi)\vphantom{\frac11}\right]^2}d\phi$$

So we got $$\displaystyle I = \int\frac{\tan^2 \phi }{\left[\cos\left(\tan \phi-\phi\right)\vphantom{\frac11}\right]^2}d\phi$$

Now Put $(\tan\phi-\phi) = t\;,$ Then $(\sec^2 \phi-1)d\phi = dt\Rightarrow \tan^2 \phi d\phi = dt$

So Integral $$\displaystyle I = \int\frac{1}{\cos^2 t}dt = \int\sec^2t dt = \tan t+\mathcal{C}=\tan(\tan \phi-\phi)+\mathcal{C}$$

So Integral $$\displaystyle I = \tan (x-\tan^{-1}x)+\mathcal{C} = \frac{\tan x-x}{1+\tan x\cdot x}+\mathcal{C} = \left[\frac{\sin x-x\cdot \cos x}{\cos x+x\cdot \sin x}\vphantom{\frac11}\right]+\mathcal{C}$$

juantheron
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