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Evaluation of Integral $\displaystyle \int\frac{1}{(2+3\sin x)^2}dx$

$\bf{My\; Try::}$ Using Integration by parts,

Let $$\displaystyle I=\int\frac{1}{(2+3\sin x)^2}dx = \int \frac{1}{\cos x}\cdot \frac{\cos x}{(2+3\sin x)^2}dx$$

$$\displaystyle I = -\frac{1}{3 \cos x}\cdot \frac{1}{(2+3\sin x)}+\frac{1}{3}\int \frac{\sin x}{\cos^2 x}\cdot \frac{1}{(2+3\sin x)}dx$$

Now Let $$\displaystyle J=\int \cdot \frac{\sin x}{(2+3\sin x)}\cdot \frac{1}{\cos^2 x} = \int \frac{\sin x}{(2+3\sin x)}\cdot \sec^2 xdx$$

again Using Integration parts

$$\displaystyle J = \frac{\sin x}{(2+3\sin x)}\cdot \tan x - \int \frac{d}{dx}\left\{\frac{\sin x}{(2+3\sin x)}\right\}\cdot \tan xdx$$

So $$\displaystyle J = \frac{\sin x}{(2+3\sin x)}\cdot \tan x-2\int\frac{\sin x}{(2+3\sin x)^2}dx$$

$$\displaystyle J = \frac{\sin x}{(2+3\sin x)}\cdot \tan x-\frac{2}{3}\int \frac{[(2\sin x+3)-3]}{(2\sin x+3)^2}dx$$

So $$\displaystyle J = \frac{\sin x}{(2+3\sin x)}\cdot \tan x- \frac{2}{3}\int\frac{1}{(2\sin x+3)}dx+4I$$

Now Let $$\displaystyle K = \int\frac{1}{2\sin x+3}dx = \int\frac{2\sin x-3}{4\sin^2 x-9}dx = -2\int\frac{\sin x}{5+4\cos^2 x}dx+3\int\frac{\sec^2 x}{5\tan^2 x+9}dx$$

My Question is can we solve it Without Using Integration by parts

If Yes, The plz explain here

thanks

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  • $\begingroup$ Maybe tangent half-angle substitution does the trick here. $\endgroup$ – Dmoreno Aug 31 '14 at 10:13
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Hint: Let $~I(b)=\displaystyle\int\frac{dx}{a\sin x+b},~$ and solve it using the Weierstrass substitution. Then notice

that your original integral is nothing more than $-I'(b)$.

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If you set $x=2\arctan\theta$, you end with: $$\frac{1}{2}\int\frac{1+\theta^2}{(1+3\theta+\theta^2)^2}\,d\theta$$ that can be computed through simple fractions decomposition.

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  • $\begingroup$ Can you please elaborate a bit ?? $\endgroup$ – lab bhattacharjee Aug 31 '14 at 13:16
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Yes, apply the Weierstrass substitution $x = 2 \arctan t$, which gives

$${\int \dfrac{1}{\left(2 + 3 \cdot \frac{2t}{1 + t^2}\right)^2} \cdot \dfrac{2\,dt}{1 + t^2} = \frac{1}{2}\int \dfrac{(1 + t^2) \,dt}{(t^2 + 3t + 1)^2}}$$.

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  • $\begingroup$ How does one produce a display-size integral symbol? $\endgroup$ – Travis Aug 31 '14 at 10:18
  • $\begingroup$ use $$..$$ ${}{}{}$ $\endgroup$ – RE60K Aug 31 '14 at 10:19

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