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Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 118 Maths Textbook Solution.

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Answer:\frac{e^{2 x} \tan x}{2}+C

Hint: to solve this equation, we have to assume 2x as t

Given:\int e^{2} x\left(\frac{1+\sin 2 x}{1+\cos 2 x}\right) d x


\int e^{2 x}\left(\frac{1+\sin 2 x}{1+\cos 2 x}\right) d x

\int e^{x} f(x)+f^{\prime}(x) d x=e^{x}(x)+C                   

\begin{aligned} &2 x=t \\ &2 d x=d t \\ &d x=\frac{d t}{2} \end{aligned}                                                             \left[\begin{array}{l} \cos ^{2} \theta=2 \cos ^{2} \theta-1 \\ 1+\cos 2 \theta=2 \cos ^{2} \theta \\ 1+\cos 2 x=2 \cos ^{2} x \\ \sin ^{2} x=2 \sin x \cos x \end{array}\right]                         

I=\int e^{t}\left(\frac{1+\sin t}{1+\cos t}\right) \frac{d t}{2}

=\int e^{t}\left(\frac{1+\sin t}{1+\cos t}\right) \cdot \frac{d t}{2}

=\frac{1}{2} \int e^{t}\left(\frac{1+2 \sin \frac{t}{2} \cdot \cos \frac{t}{2}}{2 \cos ^{2} \frac{t}{2}}\right) d t

=\frac{1}{2} \int e^{t}\left(\frac{1}{2} \sec ^{2} \frac{t}{2}+\frac{2 \sin \frac{t}{2} \cos \frac{t}{2}}{2 \cos ^{2} \frac{t}{2}}\right) d t

=\frac{1}{2} \int e^{t}\left(\frac{1}{2} \sec ^{2} \frac{t}{2}+\tan \frac{t}{2}\right) d t

=\frac{1}{2} e^{t} \tan \frac{t}{2}+C

=\frac{e^{t} \tan \frac{t}{2}}{2}+C

=\frac{e^{2 x} \tan x}{2}+C . \text { resubs. } t=2 x

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