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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.

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

Answers (1)

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

Solution:

\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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