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  • Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.10 Question 7

Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.10 Question 7

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Answer:  -\frac{2}{5}(1-x)^{\frac{5}{2}}+\frac{4}{3}(1-x)^{\frac{3}{2}}+-2(1-x)^{\frac{1}{2}}+c

Hint: Use substitution method to solve this type of integral

Given:  \int \frac{x^{2}}{\sqrt{1-x}} d x

Solution: Let  I=\int \frac{x^{2}}{\sqrt{1-x}} d x

Substitute 1-x=t \Rightarrow d x=-d t  then

\begin{aligned} &I=\int \frac{(1-t)^{2}}{\sqrt{t}}(-d t) \qquad(\because x=1-t) \\ & \end{aligned}

\Rightarrow I=-\int\left(\frac{1+t^{2}-2 t}{\sqrt{t}}\right) d t \qquad\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right]

\begin{aligned} &\Rightarrow I=-\int\left\{\frac{1}{\sqrt{t}}+\frac{t^{2}}{\sqrt{t}}-\frac{2 t}{\sqrt{t}}\right\} d t \\ & \end{aligned}

\Rightarrow I=-\int\left(t^{\frac{-1}{2}}+t^{2-\frac{1}{2}}-2 t^{1-\frac{1}{2}}\right) d t \\

\Rightarrow I=-\left[\int t^{-\frac{1}{2}} d t+\int t^{\frac{3}{2}} d t-2 \int t^{\frac{1}{2}} d t\right.

\Rightarrow I=-\left[\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\frac{t^{\frac{3}{2}+1}}{\frac{3}{2}+1}-2 \frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c\right] \qquad\left[\because \int t^{n} d t=\frac{t^{n+1}}{n+1}+c\right]

\begin{aligned} &\Rightarrow I=-\left[\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+\frac{t^{\frac{5}{2}}}{\frac{5}{2}}-2 \frac{t^{\frac{3}{2}}}{\frac{3}{2}}+c\right] \\ & \end{aligned}

\Rightarrow I=-\left[2 . t^{\frac{1}{2}}+\frac{2}{5} t^{\frac{5}{2}}-2 \cdot \frac{2}{3} t^{\frac{3}{2}}\right]+c \\

\Rightarrow I=-2(1-x)^{\frac{1}{2}}-\frac{2}{5}(1-x)^{\frac{5}{2}}+\frac{4}{3}(1-x)^{\frac{3}{2}}+c \qquad[\because 1-x=t]

\therefore I=-\frac{2}{5}(1-x)^{\frac{5}{2}}+\frac{4}{3}(1-x)^{\frac{3}{2}}+-2(1-x)^{\frac{1}{2}}+c

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