More integration

**Ozymandias** · January 13th 2012, 02:15 AM

solve

$\int \frac{3+4\cos{x}}{(4+3\cos{x})^2}dx$

**Prove It** · January 13th 2012, 02:41 AM

Originally Posted by Ozymandias

solve

$\int \frac{3+4\cos{x}}{(4+3\cos{x})^2}dx$

integral[(3 + 4*Cos[x])/((4 + 3*Cos[x])^2)] - Wolfram|Alpha

Click "Show steps".

**sbhatnagar** · January 13th 2012, 02:54 AM

$\begin{align*} \int \frac{3+4\cos{x}}{(4+3\cos{x})^2}dx &= \int \frac{3(\cos^2{x}+\sin^2{x})+4\cos{x}}{(4+3\cos{x} )^2}dx \\ &= \int \frac{3\cos^2{x}+3\sin^2{x}+4\cos{x}}{(4+3\cos{x}) ^2}dx \\ &= \int \frac{\cos{x}(3\cos{x}+4)-\sin{x}(-3\sin{x})}{(4+3\cos{x})^2}dx \\ &=\int \frac{\frac{d(\sin{x})}{dx}\cdot(3\cos{x}+4)-\sin{x}\cdot\frac{d(3\cos{x}+4)}{dx}}{(4+3\cos{x}) ^2}dx \end{align*}$

Now recall that: $\frac{f'(x)g(x)-g'(x)f(x)}{[g(x)]^2}=\frac{d \Big( \frac{f(x)}{g(x)}\Big)}{dx}$

Therefore the integral becomes: $\int \frac{d \Big( \frac{\sin{x}}{4+3\cos{x}}\Big)}{dx}dx=\frac{\sin{ x}}{4+3\cos{x}}+C$

**Ozymandias** · January 13th 2012, 04:43 AM

Thank You guyz, for your precious help...

**jacks** · January 13th 2012, 11:13 PM

$\displaystyle \int \frac{3+4\cos x}{(4+3\cos x)^2}dx$

divide both $N_{r}$ and $D_{r}$ by $\sin^2 x$

$\displaystyle \int \frac{3\csc^2 x+4\csc x.\cot x}{4\csc x+3\cot x}dx$

Now Put $4\csc x+3\cot x =t \Leftrightarrow -(4\csc x.\cot x+3\csc^2 x)dx = dt$

So $\displaystyle -\int \frac{1}{t^2}dt = \frac{1}{t}+C$

So $\displaystyle \int \frac{3+4\cos x}{(4+3\cos x)^2}dx = \frac{1}{4\csc x+3\cot x}+C$

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