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

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

Answers (1)

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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