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Need solution for RD Sharma maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.29 Question 12 maths textbook solution.         

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Answer : I=-\frac{1}{3}\left(3-4 x-x^{2}\right)^{\frac{3}{2}}+\frac{1}{2}(x+2) \sqrt{3-4 x-x^{2}}+\frac{7}{2} \sin ^{-1}\left(\frac{x+2}{\sqrt{7}}\right)+C

Hint: To solve the given integration,  we express the linear term as a derivative of quadratic into constant plus another constant

Given : \int(x+3) \sqrt{3-4 x-x^{2}} d x

Solution :

\begin{aligned} &I=\int(x+2+1) \sqrt{3-4 x-x^{2}} d x \\ &I=\int(x+2) \sqrt{3-4 x-x^{2}} d x+\int \sqrt{3-4 x-x^{2}} d x \\ &\text { Let }, 3-4 x-x^{2}=u^{2} \end{aligned}

\begin{aligned} &\Rightarrow(-4-2 x) d x=2 u d u \\ &\Rightarrow-2(x+2) d x=2 u d u \\ &\Rightarrow(x+2) d x=-u d u \end{aligned}

\begin{aligned} &I=-\int u \sqrt{u^{2}} d u+\int \sqrt{-\left(x^{2}+4 x+4-4\right)+3} d x \\ &I=-\int u^{2} d u+\int \sqrt{-(x+2)^{2}+7} d x \end{aligned}

Use the formula : \int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+C

And \left[\int x^{n} d x=\frac{x^{n}+1}{n+1}+C\right]

\begin{aligned} &I=-\frac{u^{3}}{3}+\int \sqrt{(\sqrt{7})^{2}-(x+2)^{2}} d x \\ &I=\frac{-\left(3-4 x-x^{2}\right)^{3 / 2}}{3}+\frac{x+2}{2} \sqrt{3-4 x-x^{2}}+\frac{7}{2} \sin ^{-1}\left(\frac{x+2}{\sqrt{7}}\right)+C \end{aligned}I=-\frac{1}{3}\left(3-4 x-x^{2}\right)^{\frac{3}{2}}+\frac{1}{2}(x+2) \sqrt{3-4 x-x^{2}}+\frac{7}{2} \sin ^{-1}\left(\frac{x+2}{\sqrt{7}}\right)+C

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