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  • Provide solution for RD Sharma Maths Class12 Chapter 21 Differential Equation exercise 21.5 question 10.

Provide solution for RD Sharma Maths Class12 Chapter 21 Differential Equation exercise 21.5 question 10.

Answers (1)

Answer: y=\frac{\sin^3x}{3}-\frac{\sin^5x}{5}+\frac{(2x+1)^{\frac{5}{2}}}{10}-\frac{(2x+1)^{\frac{3}{2}}}{3}

Hint: You have to integrate by applying integration of \cos^n x and \sin^n x.

Given: \frac{d y}{d x}=\cos ^{3} x \sin ^{2} x+x \sqrt{2 x+1}

Solution:\frac{d y}{d x}=\cos ^{3} x \sin ^{2} x+x \sqrt{2 x+1}

dy=(\cos ^{3} x \sin ^{2} x+x \sqrt{2 x+1})dx

Integrating both sides,

\begin{aligned} & \int d y=\int \cos ^{3} x \sin ^{2} x d x+\int x \sqrt{2 x+1} d x \\ =& \int \cos x \times\left(\cos ^{2} x\right) \sin ^{2} x d x+\int x \sqrt{2 x+1} d x \\ =& \int \cos x\left(1-\sin ^{2} x\right) \sin ^{2} x d x+\int x \sqrt{2 x+1} d x \end{aligned}

Put \sin x=t

\cos x \; dx=dt

\begin{aligned} &y=\int\left(1-t^{2}\right) t^{2} d t+\frac{1}{2}(2 x+1-1 \sqrt{2 x+1} d x) \\ &y=\left(t^{2}-t^{4}\right) d t+\frac{1}{2} \int(2 x+1) \cdot(\sqrt{2 x+1}) d x-\int \sqrt{2 x+1} d x \\ &y=\frac{t^{3}}{3}+\frac{t^{5}}{5}+\frac{1}{2} \int(2 x+1)^{\frac{3}{2}} d x-\int \sqrt{2 x+1} d x \\ &y=\frac{\left(\sin ^{3} x\right)^{3}}{3}+\frac{\left(\sin ^{5} x\right)^{5}}{5}+\frac{1}{2} \frac{(2 x+1)^{\frac{3}{2}}}{\frac{8}{2}+1}-\frac{(2 x+1)^{\frac{1}{2}}}{\frac{1}{2}+1}+C \\ &y=\frac{\sin ^{3} x}{3}-\frac{\sin ^{5} x}{5}+\frac{(2 x+1)^{\frac{5}{2}}}{10}-\frac{(2 x+1)^{\frac{3}{2}}}{3} \end{aligned}

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