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We are to solve $$\int \frac1{(3+4\sin x)^2}\,dx.$$

I had tried expanding the denominator and substituting $\sin x$ in terms of $\tan(x/2)$ and then putting $\tan(x/2) =t$. But, this made the integration even more complex. I got a large bi-quadratic equation in the denominator.

If I am able to get $\cos x$ in the numerator ,then the question can be proceeded from there.

Any ideas or other methods by which I could solve?

Thanks in advance.

StubbornAtom
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Amritansh Singhal
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3 Answers3

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If we put $3 + 4 \sin(x) = u$, we get:

$$I = \int\frac{du}{u^2\sqrt{7 +6 u - u^2}}$$

Substituting $u = \frac{1}{t}$ yields:

$$I = -\int\frac{tdt}{\sqrt{7t^2 +6 t - 1}} = -\frac{1}{14}\int\frac{(14t + 6)dt}{\sqrt{7t^2 +6 t - 1}} +\frac{3}{7}\int\frac{dt}{\sqrt{7t^2 +6 t - 1}}$$

The first term on the r.h.s. yields $-\frac{1}{7}\sqrt{7t^2 + 6 t -1}$, while the second term is easily evaluated by writing the argument of the square root in terms of a perfect square and then doing a hyperbolic substitution.

Count Iblis
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Let us try this manually. As in usual first we can try the substitution $$\tan\left(\frac x2\right)=t$$ which gives $$dx=\dfrac{2dt}{1+t^2}$$ and $$\sin x=\dfrac{2t}{1+t^2}.$$ Then our integral reduces to $$\int \frac1{(3+4\sin x)^2}\,dx=2\int \dfrac{t^2+1}{(3t^2+8t+3)^2}\,dt.$$ Note that the fact $\dfrac{d}{dx}\Big(\dfrac uv\Big)=\dfrac{vu'-uv'}{v^2},$ lets us to assume our integral should consists some thing of the form $$\dfrac{f(t)}{3t^2+8t+3},$$ where $f$ is a polynomial whose degree is at most $2.$ By inspection we can see that $$\color{Green}{\dfrac{d}{dx}\Big(\dfrac{t^2+3t+1}{3t^2+8t+3}\Big)=\dfrac{-t^2+1}{(3t^2+8t+3)^2}}.$$

Therefore $$\int \dfrac{t^2+1}{(3t^2+8t+3)^2}\,dt=-\Big(\dfrac{t^2+3t+1}{3t^2+8t+3}\Big)+2\color{Red}{\int \dfrac{1}{(3t^2+8t+3)^2}\,dt}.$$ For evaluate the red colored integral, find partial factions and continue as we usually do...

Good Luck with your problem.

Bumblebee
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HINT: $$\displaystyle \int \frac{dx}{(a+4sinx)^2} = -\frac{d}{da}\int \frac{dx}{a+4sinx}$$ $$\int \frac{dx}{a+4sinx}= \frac{2arctan\left(\frac{atan(\frac{x}{2})+4}{\sqrt{a^2-16}}\right)}{\sqrt{a^2 -16}}$$

Meadara
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  • @meadra Thanks for attempting. But would there be a cosx in the numerator in the first step you wrote? – Amritansh Singhal Aug 30 '16 at 23:42
  • No since we are taking $a$ as a parameter,(that is what makes this technique useful for higher powers of the integral) eventhough in this case the best method is what Nilan used (tan sub). – Meadara Aug 31 '16 at 21:13