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$$\int \frac{dx}{(3+2 \sin x)^2}$$

ATTEMPT:-

Re Writing the integral as:

$I=\int \frac{2\cos x \sec x \,dx}{2(3+2 \sin x)^2}$ and using by parts:-

$\int u\,dv= uv-\int v\,du$

Here $u=\sec x \implies du=\sec x \tan x \,dx$

$\quad$ $dv=\frac{2\cos x \,dx}{2(3+2sinx)^2} \implies v=\frac{-1}{2(3+2\sin x)}$

$I=\frac{-\sec x\,dx}{2(3+2\sin x)} +\int \frac{\sec x \tan x \,dx}{2(3+2\sin x)}$

Let $I'=\int \frac{\sec x \tan x \,dx}{2(3+2\sin x)}$

$\implies I'=\int \frac{\sin x \,dx}{2\cos^2x(3+2\sin x)}$

$\implies I'=\int \frac{\sin \,dx}{2(1-\sin^2x)(3+2\sin x)}$

$\implies I'=\int \frac{-dx}{4(1+\sin x)} +\frac{3dx}{5(2\sin x+3)} + \frac{-dx}{20(\sin x-1)}$ which can be easily done by weierstrass's substitution.

But I am not able to modify my answer as given in the text.

Text Ans:-$\frac{2\cos x\,dx}{5(3+2\sin x)^2} + \frac{2}{5\sqrt{5}} \arctan\frac{(3\tan(\frac{x}{2})+2)}{\sqrt{5}} +c.$

My Ans:-$\frac{-\sec x\,dx}{2(3+2\sin x)} + \frac{6}{5\sqrt{5}} \arctan\frac{(3\tan(\frac{x}{2})+2)}{\sqrt{5}}-10\tan x+15\sec x-15.$

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    $\begingroup$ Why not simply apply the Weierstraß substitution to the original integrand? $\endgroup$ – Travis Willse Nov 3 '15 at 21:45
  • $\begingroup$ I get that the Weierstraß substitution yields $$2\int \frac{(t^2 + 1) \, dt}{(3 t + 4 t + 3)^2}$$, and this can be handled with partial fractions, a standard quadratic $u$-substitution and the fact that $\frac{d}{dv} \arctan v = \frac{1}{1 + v^2}$. $\endgroup$ – Travis Willse Nov 3 '15 at 23:08
  • $\begingroup$ @Travis it is $(3t^2+4t+3)^2$ $\endgroup$ – yasir Nov 3 '15 at 23:35
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Let $\displaystyle I = \int\frac{1}{(3+2\sin x)^2}dx\;,$ Now Put $\displaystyle \frac{2+3\sin x}{(3+2\sin x)} = t\;,$

Then $$\displaystyle \frac{(3+2\sin x)\cdot 3\cos x-(2+3\sin x)\cdot 2\cos x}{(3+2\sin x)^2}dx = dt$$

so we get $$\frac{5\cos x}{(3+2\sin x)^2}dx = dt\Rightarrow \frac{1}{(3+2\sin x)^2}dx = \frac{1}{5\cos x}dt.$$

So We get $$I = \frac{1}{5}\int \frac{1}{\cos x}dt$$

Now Above we have $$\frac{2+3\sin x}{3+2\sin x}=t\Rightarrow \sin x = \frac{2-3t}{2t-3}.$$

So we get $$\cos x= \sqrt{1-\sin^2 x} = \frac{\sqrt{5}\cdot \sqrt{1-t^2}}{2t-3}.$$

So we get $$I = \frac{1}{5\sqrt{5}}\int\frac{(2t-3)}{\sqrt{1-t^2}}dt = \frac{2}{5\sqrt{5}}\int\frac{t}{\sqrt{1-t^2}}dt-\frac{3}{5\sqrt{5}}\int\frac{1}{\sqrt{1-t^2}}dt$$

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  • $\begingroup$ How do you got that "Substitution" ? Perfect +1. $\endgroup$ – yasir Nov 5 '15 at 11:03
  • $\begingroup$ @yasir $u=\tfrac{1}{3+2\sin x}$ would be more obvious, but the answer uses a linear transformation thereof to tidy up the work with surds. $\endgroup$ – J.G. Feb 13 at 19:33
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Note $\sin x=\cos (\frac\pi2-x)=\frac{1-\tan^2(\frac\pi4-\frac x2)}{1+\tan^2(\frac\pi4-\frac x2)}$ $$\int \frac{dx}{3+2\sin x}=-2\int \frac{d(\tan(\frac\pi4-\frac x2))}{\tan^2(\frac\pi4-\frac x2) +5}=-\frac2{\sqrt{5}}\tan^{-1} \frac{\tan (\frac \pi4-\frac x2) }{ \sqrt{5}} $$ $$\left( \frac{2\cos x}{3+2\sin x}\right)’ = -\frac{3}{3+2\sin x}+\frac{5}{(3+2\sin x)^2} $$ Integrate both sides to obtain

\begin{align} I=\int \frac{dx}{(3+2\sin x)^2}=\frac 2{5}\frac{\cos x}{3+2\sin x}- \frac{6}{5\sqrt5}\tan^{-1} \frac{\tan (\frac\pi4-\frac x2 )}{ \sqrt{5}}+C \end{align}

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