# Partial Fraction of $\int \frac{ \left( \cos x + \sin 2x \right) \ \mathrm{d}x}{( 2 - \cos^2 x)(\sin x)}$

If

$$\int\frac{ \left( \cos x + \sin 2x \right) \ \mathrm{d}x}{( 2 - \cos^2 x)(\sin x)} = \int \frac{A \ \mathrm{d}x}{(\sin x)} + B \int\frac{\sin x \ \mathrm{d}x}{ 1 + \sin^2 x} + C \int \frac{\mathrm{d}x}{1 + \sin^2 x}$$

then, which of the following is correct?

1. $$A+B+C=4$$
2. $$A+B+C=2$$
3. $$A+BC=1$$
4. $$A+B+C=5$$

My approach is as follows:

$$\int\frac{\left( \cos x + \sin 2x \right) \ \mathrm{d}x}{(2 - \cos^2 x)(\sin x)}= \int \frac{\left( \cos x + \sin 2x \right) \ \mathrm{d}x}{(1 + \sin^2 x)(\sin x)} =$$

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

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

$$= \int\frac{- \sin \cos x \ \mathrm{d}x}{ 1 + \sin^2 x } + \int\frac{\cos x \ \mathrm{d}x}{ \sin x } + \int \frac{2 \cos x \ \mathrm{d}x}{1 + \sin^2 x}$$

After this step I am not able to approach

• The question is correct i cross checked it – Samar Imam Zaidi Oct 18 '18 at 8:11

The integrand can be written $$\frac{\cos x+\sin2x}{(1+\sin^2x)\sin x}$$ Let's try and determine $$A$$, $$B$$ and $$C$$ so that $$\frac{A}{\sin x}+\frac{B\sin x}{1+\sin^2x}+\frac{C}{1+\sin^2x}$$ is the same. Removing the denominators this forces $$A(1+\sin^2x)+B\sin^2x+C\sin x=\cos x+\sin2x$$ If we set $$x=0$$, we get $$A=1$$; with $$x=\pi/2$$, we get $$2A+B+C=0$$; with $$x=-\pi/2$$ we get $$2A+B-C=0$$.

The linear system $$\begin{cases} A=1\\ 2A+B+C=0\\ 2A+B-C=0 \end{cases}$$ has the solution $$A=1$$, $$B=-2$$, $$C=0$$.

Hence $$A+BC=1$$. However, the two functions are not equal as you can check at $$\pi/4$$.

In order to compute the integral, one has rather to find $$A$$, $$B$$ and $$C$$ such that $$\frac{1+2\sin x}{(1+\sin^2x)\sin x}= \frac{A}{\sin x}+\frac{B\sin x}{1+\sin^2x}+\frac{C}{1+\sin^2x}$$ which is possible and equivalent to solving for partial fractions $$\frac{1+2u}{u(1+u^2)}=\frac{A}{u}+\frac{Bu+C}{1+u^2}$$ This certainly has a solution and allows to compute your integral with $$u=\sin x$$.

This translates into $$A+Au^2+Bu^2+Cu=1+2u$$, so $$\begin{cases} A+B=0\\ C=2\\ A=1 \end{cases}$$ so $$A=1$$, $$B=-1$$ and $$C=2$$.

With this fix, the right answer would be $$A+B+C=2$$.

Write your Integrand in the form

$$\frac {\csc(x)(\cos(x)+2\sin(x)\cos(x))}{2-\cos^2(x)}$$ and then multiply numerator and denominator by

$$\frac{csc(x)(\cos(x)+2\sin(x)\cos(x))}{2-\cos^2(x)}$$ and we get

$$\newcommand{\dx}{\; \mathrm{d}}x\int \frac{\cot(x)(\csc(x)+2)\csc(x)}{\csc^2(x)+1}\dx$$

now Substitute

$$u=\csc(x)$$ then

$$du=-\cot(x)\csc(x)dx$$

and we obtain

-$$\int\frac{u+2}{u^2+1}du$$

Can you finish?