#### Provide Solution For R.D. Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Revision Exercise Question 129 Maths Textbook Solution.

Answer: $\frac{1}{2} \cot 2 x \cdot e^{2 x}+C$

Hint:Using $\int e^{x}\left(f(x)-f^{\prime}(x)\right) d x$

Given: $\int \frac{\sin 4 x-2}{1-\cos 4 x} \cdot e^{2 x} d x$

Explanation:

Let

\begin{aligned} &I=\int \frac{\sin 4 x-2}{1-\cos 4 x} \cdot e^{2 x} d x \\ &=\frac{1}{2} \int\left(\frac{\sin 2 t-2}{1-\cos 2 t}\right) e^{t} d t \\ &=\frac{1}{2} \int e^{t}\left(\frac{\sin t \cos t-1}{\sin ^{2} t}\right) d t \end{aligned}                                    $\left[\begin{array}{l} \text { put } 2 x=t \\ 2 d x=d t \\ d x=\frac{1}{2} d t \end{array}\right]$

\begin{aligned} &=\frac{1}{2} \int e^{t}\left(\frac{\cos t}{\sin t}-\frac{1}{\sin ^{2} t}\right) d t \\ &=\frac{1}{2} \int e^{t}\left(\cot t-\cos e c^{2} t\right) d t \\ &=\frac{1}{2} \int e^{t} \cot d t-\frac{1}{2} \int e^{t} \cos e c^{2} t d t \\ &=\frac{1}{2}\left[\cot \cdot e^{t}-\int\left(-\cos e c^{2} t\right) \cdot e^{t} d t-\frac{1}{2} \int e^{t} \cos e c^{2} t d t\right] \\ &=\frac{1}{2} \cot t \cdot e^{t}+\frac{1}{2} \int \cos e c^{2} t \cdot e^{t} d t-\frac{1}{2} \int e^{t} \cos e c^{2} t d t \\ &=\frac{1}{2} \cot t \cdot e^{t}+C \\ &=\frac{1}{2} \cot 2 x e^{2 x}+C \end{aligned}