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  • provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.9 question 14

provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise  18.9 question 14

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

Hint: Use substitution method to solve this integral.

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

Solution:

        \begin{aligned} &\text { Let } I=\int \frac{\sin ^{3} x}{\sqrt{\cos x}} d x=\int \frac{\sin ^{2} x \cdot \sin x}{\sqrt{\cos x}} d x \\ &=\int \frac{\left(1-\cos ^{2} x\right) \cdot \sin x}{\sqrt{\cos x}} d x \quad\left[\begin{array}{l} \because \sin ^{2} x+\cos ^{2} x=1 \\ \Rightarrow \sin ^{2} x=1-\cos ^{2} x \end{array}\right] \\ &\text { Put } \cos x=t \Rightarrow-\sin x d x=d t \Rightarrow \sin 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-t^{2}}{\sqrt{t}}\right\} d t \\ &=-\int\left\{t^{\frac{-1}{2}}-t^{2-\frac{1}{2}}\right\} d t=-\int\left\{t^{\frac{-1}{2}}-t^{\frac{4-1}{2}}\right\} d t \\ &=-\int\left\{t^{\frac{-1}{2}}-t^{\frac{3}{2}}\right\} 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{\cos x}+\frac{2}{5}(\cos x)^{\frac{5}{2}}+c\; \; \; \; \; [\because t=\cos x] \end{aligned}

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