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Please solve RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 13 maths textbook solution

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Answer: 2 \sqrt{\sin x}-\frac{2}{5}(\sin x)^{\frac{5}{2}}+c

Hint: Use substitution method to solve this integral.

Given: \int \frac{\cos ^{3} x}{\sqrt{\sin x}} d x


        \begin{aligned} &\text { Let } I=\int \frac{\cos ^{3} x}{\sqrt{\sin x}} d x=\int \frac{\cos ^{2} x \cdot \cos x}{\sqrt{\sin x}} d x \\ &=\int \frac{\left(1-\sin ^{2} x\right) \cdot \cos x}{\sqrt{\sin x}} d x \quad\left[\begin{array}{l} \because \sin ^{2} x+\cos ^{2} x=1 \\ \Rightarrow \cos ^{2} x=1-\sin ^{2} x \end{array}\right] \\ &\operatorname{Put} \sin x=t \Rightarrow \cos x d x=d t \text { then } \end{aligned}

        \begin{aligned} I &=\int \frac{\left(1-t^{2}\right)}{\sqrt{t}} d t=\int\left\{\frac{1}{\sqrt{t}}-\frac{t^{2}}{\sqrt{t}}\right\} d t \\ &=\int t^{\frac{-1}{2}} d t-\int t^{2-\frac{1}{2}} d t=\int t^{\frac{-1}{2}} d t-\int t^{\frac{4-1}{2}} d t \\ &=\int t^{\frac{-1}{2}} d t-\int t^{\frac{3}{2}} d t \end{aligned}

            =\frac{t^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}-\frac{t^{\frac{3}{2}+1}}{\frac{3}{2}+1}+\mathrm{c} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]


            \begin{aligned} &=2 t^{\frac{1}{2}}-\frac{2}{5} t^{\frac{5}{2}}+c \\ &=2 \sqrt{\sin x}-\frac{2}{5}(\sin x)^{\frac{5}{2}}+c\; \; \; \; [\because t=\sin x] \end{aligned}

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