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

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

Solution:

        \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]

            =\frac{t^{\frac{1}{2}}}{\frac{1}{2}}-\frac{t^{\frac{5}{2}}}{\frac{5}{2}}+\mathrm{c}

            \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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