Question

$\frac{dy}{dx}={\sin }^{3}x{\cos }^{4}x+x\sqrt{x+1}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}={\sin }^{3}x{\cos }^{4}x+x\sqrt{x+1} \\ \Rightarrow dy=\left({\sin }^{3}x{\cos }^{4}x+x\sqrt{x+1}\right)dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int dy=\int \left({\sin }^{3}x{\cos }^{4}x+x\sqrt{x+1}\right)dx \\ \Rightarrow y=\int {\sin }^{3}x{\cos }^{4}xdx+\int x\sqrt{x+1}dx \\ \Rightarrow y=I_1+I_2.....\left(1\right) \\ \mathrm{Here}, \\ {I}_1=\int {\sin }^{3}x{\cos }^{4}xdx \\ {I}_1=\int x\sqrt{x+1}dx \\ \mathrm{Now}, \\ {I}_1=\int {\sin }^{3}x{\cos }^{4}xdx \\ =\int \left(1-{\cos }^{2}x\right){\cos }^{4}x\sin xdx \\ \mathrm{Putting}t=\cos x,\mathrm{we}\mathrm{get} \\ dt=-\sin xdx \\ \therefore I_1=-\int t^4\left(1-t^2\right)dt \\ =\int \left(t^6-t^4\right)dt \\ =\frac{t^7}{7}-\frac{t^5}{5}+C_1 \\ =\frac{{\cos }^{7}x}{7}-\frac{{\cos }^{5}x}{5}+C_1 \\ {I}_2=\int x\sqrt{x+1}dx \\ \mathrm{Putting}{t}^2=x+1,\mathrm{we}\mathrm{get} \\ 2tdt=dx \\ \therefore I_2=2\int \left(t^2-1\right)t^{2}dt \\ =2\int \left(t^4-t^2\right)dt \\ =\frac{2t^5}{5}-\frac{2t^3}{3}+C_2 \\ =\frac{2{\left(x+1\right)}^{{\displaystyle \frac{5}{2}}}}{5}-\frac{2{\left(x+1\right)}^{\frac{3}{2}}}{3}+C_2 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}{I}_1\mathrm{and}{I}_2\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ y=\frac{{\cos }^{7}x}{7}-\frac{{\cos }^{5}x}{5}+C_1+\frac{2{\left(x+1\right)}^{{\displaystyle \frac{5}{2}}}}{5}-\frac{2{\left(x+1\right)}^{\frac{3}{2}}}{3}+C_2 \\ y=\frac{{\cos }^{7}x}{7}-\frac{{\cos }^{5}x}{5}+\frac{2{\left(x+1\right)}^{{\displaystyle \frac{5}{2}}}}{5}-\frac{2{\left(x+1\right)}^{\frac{3}{2}}}{3}+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left[\because C=C_1\mathit{+}{C}_2\right] \\ \mathrm{Hence},y=\frac{{\cos }^{7}x}{7}-\frac{{\cos }^{5}x}{5}+\frac{2{\left(x+1\right)}^{{\displaystyle \frac{5}{2}}}}{5}-\frac{2{\left(x+1\right)}^{\frac{3}{2}}}{3}+\mathrm{C}\mathrm{is}\mathrm{the}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}. \end{gathered}$$