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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

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  • 1
    $\begingroup$ The question is correct i cross checked it $\endgroup$ – Samar Imam Zaidi Oct 18 '18 at 8:11
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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 $A+B+C=2$.

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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?

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