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  • Need Solution For RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.23 Question 11 Maths Textbook Solution.

Need Solution For  RD Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.23 Question 11 Maths Textbook Solution.

Answers (1)

Answer : \sqrt{2} \tan ^{-1}\left(\frac{\tan \left(\frac{x}{2}\right)+1}{\sqrt{2}}\right)+c

Hint : To solve this equation we have to convert sinx and cosx in terms of tanx

Given : \int \frac{1}{2+\sin x+\cos x} d x

Solution : We use the formula : \left[\begin{array}{l} \sin x=\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}} \\ \cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}} \end{array}\right]

\begin{aligned} &=\int \frac{1}{2+\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}+\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}} d x\\ &=\int \frac{1+\tan ^{2} \frac{x}{2}}{2+2 \tan ^{2} \frac{x}{2}+2 \tan \frac{x}{2}+1-\tan ^{2} \frac{x}{2}} d x \end{aligned}

Multiple 2 by numerators and denominator

\int \frac{2\left(\sec ^{2} \frac{x}{2}\right)}{2\left(\tan ^{2} \frac{x}{2}+2 \tan \frac{x}{2}+3\right)} d x

\begin{aligned} &t=\tan \frac{x}{2} \\ &\frac{d t}{d x}=\frac{1}{2} \sec ^{2} \frac{x}{2} \\ &d t=\frac{1}{2} \sec ^{2} \frac{x}{2} d x \end{aligned}

\begin{aligned} &=2 \int \frac{d t}{t^{2}+2 t+3} \\ &=2 \int \frac{d t}{\left(t^{2}+2 t+1\right)+2} \\ &=2 \int \frac{d t}{(t+1)^{2}+2} \end{aligned}

\begin{aligned} &t+1=u \\ &1=\frac{d u}{d t} \\ &d u=d t \\ &=2 \int \frac{d u}{(u)^{2}+(\sqrt{2})^{2}} \end{aligned}

\begin{aligned} &\int \frac{d x}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c \\ &=\sqrt{2} \times \frac{\sqrt{2}}{\sqrt{2}} \tan ^{-1}\left(\frac{u}{\sqrt{2}}\right)+c \\ &=\sqrt{2} \tan ^{-1}\left(\frac{t+1}{\sqrt{2}}\right)+c \end{aligned}

=\sqrt{2} \tan ^{-1}\left(\frac{\tan \frac{x}{2}+1}{\sqrt{2}}\right)+c

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