Integrate $\int \frac{x\cos x}{\sin^2x}dx$

Question

$$\int \frac{x\cos x}{\sin^2x}dx$$

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

How can I find the two fractions? if there are at all?

Answer 1

Note that $$\frac{\cos x}{\sin^2 x}=\csc x \cot x$$ Applying integration by parts and differentiating 'x' we get,

$$I= \int \frac{x\cos x}{\sin^2x}dx=-x \csc x +\int \csc x\,dx$$

Now can you proceed?

Answer 2

$$\int \frac{x\cos x}{\sin^2x}dx$$ Let $\sin x=u$. Then, $\cos x dx=du$ and $x=\arcsin u$. $$\int \frac{x\cos x}{\sin^2x}dx=\int \frac{\arcsin u}{u^2}du$$ This can be integrated by parts taking $\arcsin u$ as the first function and $\frac{1}{u^2}$ as the second function.

Answer 3

$$\int \frac{x\cos x}{(1-\cos x)(1+\cos x)}dx=\int\frac{x \cos x}{1-\cos^2 x}dx=\int x\cot x \csc xdx$$ And now by parts: $f=x$ and $df=dx$ and $dg=\cot x \csc x dx$ and $g=-\csc$

$$=-x \csc x+\int\csc x dx$$

Multiply numenator and denominator of $\csc x$ by $\cot x +\csc x$

$$-x\csc x-\int\frac{-\cot x \csc x-\csc^2 x}{\cot x+\csc x}dx$$

And now set $t=\cot x +\csc x$ and $dt=(-\csc^2 x-\cot x \csc x)dx$

$$=-x\csc x-\int \frac 1 t dt$$ Can you finish?

Answer 4

For the calculation of $$I=\int \csc(x)\,dx$$ the tangent half angle subsitution is very useful and makes everything simple $$t=\tan(\frac x2)\implies dx=\frac {2dt}{1+t^2}\qquad \csc(x)=\frac{1+t^2}{2t}$$ All of this makes $$I=\int \frac{dt} t$$

Answer 5

Since you wanted to solve using partial fractions, $$\int\frac{x\cos x}{(1+\cos x)(1-\cos x)}dx=\frac{1}{2}\int\frac{x(1+\cos x)-x(1-\cos x)}{(1+\cos x)(1-\cos x)}dx$$ $$=\frac{1}{2}\int\frac{x}{1-\cos x}dx- \frac{1}{2}\int{\frac{x}{1+\cos x}}dx$$ $$= \frac{1}{2}\int \frac{x}{2\sin^2\frac{x}{2}}dx- \frac{1}{2}\int \frac{x}{2\cos^2\frac{x}{2}}dx$$ $$= \frac{1}{2}\int x\sec^2\frac{x}{2}dx-\frac{1}{2}\int x\csc^2\frac{x}{2}dx$$ Now use integration by parts. I will solve one term $$\int x\sec^2\frac{x}{2}dx=2x\tan\frac{x}{2}-2\int\tan\frac{x}{2}dx$$

Answer 6

Notice, using integration by parts $$\int \frac{x\cos x}{\sin^2 x}\ dx$$ $$=x\int \frac{\cos x}{\sin^2 x}\ dx-\int \left(\frac{d}{dx}(x)\cdot \int \frac{\cos x}{\sin^2 x}\ dx\right)\ dx$$ $$=x\int \frac{d(\sin x)}{\sin^2 x}-\int \left(1\cdot \int \frac{d(\sin x)}{\sin^2 x}\ dx\right)\ dx$$ $$=x\cdot \frac{-1}{\sin x}-\int \left(\frac{-1}{\sin x}\right)\ dx$$ $$=-x\csc x +\int \csc x\ dx$$ $$=-x\csc x +\ln|\csc x-\cot x|+C$$

Answer 7

You should recognize the derivative of the cosecant. Then by parts,

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

The last integral is easily solved with

$$\int\frac{dx}{\sin(x)}=\int\frac{\sin(x)}{\sin^2(x)}dx=\int\frac{\sin(x)}{1-\cos^2(x)}dx=-\text{artanh}(\cos(x)).$$