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$$\int \frac{x^2+(n(n-1))}{(x\sin x +n\cos x)^2 } dx$$ I know this is an homework problem, but I really couldn't think of any way to solve it. Like DI Method (No go) , What kind of substitution as denominator is trigonometric whereas Numerator is algebric. Thought of n(n-1) can come by double differentiating but.. like how would we have it here ... etc confusing and weird thoughts. Please help me out

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  • $\begingroup$ It is $$\frac{n \sin (x)-x \cos (x)}{n \cos (x)+x \sin (x)}+C$$ $\endgroup$ – Dr. Sonnhard Graubner May 19 at 6:10
  • $\begingroup$ @Dr.SonnhardGraubner How ?? Please show /tell the working and thought process $\endgroup$ – user232243 May 19 at 6:16
  • $\begingroup$ Integration by parts might help, after some manipulation. $\endgroup$ – arya_stark Jun 19 at 8:28
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$$I=\int\frac{x^2+n(n-1)}{(x\sin x+n\cos x)^2}dx$$

Put $x=n\tan \theta\;\;dx=n\sec^2\theta d\theta$

$$I=\int\frac{n^2\tan^2\theta+n^2-n}{(n\tan\theta\sin(n\tan \theta)+n\cos(n\tan \theta))^2}\cdot n\sec^2\theta d\theta$$

$$I=\int\frac{n\sec^2(\theta)-1}{\cos^2(n\tan\theta-\theta)}d\theta.$$

Put $n\tan \theta-\theta=u$ and $(n\sec^2\theta-1)d\theta=du$

$$I=\int\frac{1}{\cos^2u}du=\int\sec^2(u)du=\tan u+C$$

$$I=\tan(n\tan\theta-\theta)+C=\tan\bigg(x-\tan^{-1}\frac{x}{n}\bigg)+C$$

$$I=\frac{n\sin x-x\cos x}{x\sin x+n\cos x}+C$$

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  • $\begingroup$ Thank You so much. If you don't mind me asking , what was your thought process , behind it that you could think of substitution of $x=n*tan(\theta)$ $\endgroup$ – user232243 May 20 at 3:44
  • $\begingroup$ From Dr. Answer...... $\endgroup$ – jacky May 20 at 8:29
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$I = \int\frac{x^2 + n(n-1)}{(x\sin x + n\cos x )^2}dx$

Now we'll try to convert it into the form of $\frac{a}{y} + \frac{b}{y^2}$, where $a,b$ are functions of $x$ and $y$ is the denominator.

$$x^2+n(n-1) =(x\sin x + (n-1)\cos x)(x\sin x+n\cos x)-((1-n)\sin x + x\cos x)(n\sin x - x\cos x)$$

$$$$

$I = \int \big[\frac{(x\sin x + (n-1)\cos x)(x\sin x+n\cos x)}{(x\sin x + n\cos x )^2} - \frac{((1-n)\sin x + x\cos x)(n\sin x - x\cos x)}{(x\sin x + n\cos x )^2}\big]dx$

Now, $I = \int\big[\frac{(x\sin x + (n-1)\cos x)}{(x\sin x + n\cos x )} - \frac{((1-n)\sin x + x\cos x)(n\sin x - x\cos x)}{(x\sin x + n\cos x )^2}\big]dx$

Let $I_1 = \int\frac{(x\sin x + (n-1)\cos x)}{(x\sin x + n\cos x )}dx$ , $I_2 = \frac{((1-n)\sin x + x\cos x)(n\sin x - x\cos x)}{(x\sin x + n\cos x )^2}dx$


In $I_2$,

let $u = n\sin x - x\cos x$, $dv = \frac{(1-n)\sin x + x\cos x}{(x\sin x + n\cos x )^2}dx$

$du = (n\cos x - \cos x + x\sin x)dx$ ,

[In $v$ , $t = x\sin x + n\cos x$, $dt = (x\cos x + \sin x - n\sin x )dx = (x\cos x + (1-n)\sin x)dx$]

$v = \int\frac{(1-n)\sin x + x\cos x}{(x\sin x + n\cos x )^2}dx = \int\frac{dt}{t^2} = -\frac{1}{t} = - \frac{1}{x\sin x + n\cos x}$


So,

$I_2 = uv - \int vdu = -(n\sin x - x\cos x)\frac{1}{x\sin x + n\cos x} + \int (n\cos x - \cos x + x\sin x).\frac{1}{x\sin x + n\cos x}dx + c$

$I_2 = - \frac{n\sin x - x\cos x}{x\sin x + n\cos x} + \int \frac{x\sin x + (n-1)\cos x}{x\sin x + n\cos x}dx +c$

$I_2 = - \frac{n\sin x - x\cos x}{x\sin x + n\cos x} + I_1 +c $

$$I = I_1 - I_2 = \frac{n\sin x - x\cos x}{x\sin x + n\cos x} + k$$

($k = -c$)

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