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


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

  • 4
    $\begingroup$ 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

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

  • $\begingroup$ 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

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}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.