Question

Solve: $\frac{dy}{dx}={\cos }^{3}x{\sin }^{2}x+x\sqrt{2x+1}$

Answer 1

$\frac{dy}{dx}={\cos }^3{\sin }^2+\sqrt[x]{x2x+1}$

$\Rightarrow dy=\left({\cos }^3{\sin }^2\right)dx+\left(\sqrt[x]{x2x+1}\right)dx$

$\Rightarrow \int dy=\int \cos .{\cos }^2{\sin }^{2}dx+\int \sqrt[x]{x2x+1}dx$

$=\int {\sin }^2(1-{\sin }^2)\cos dx+\int \sqrt[x]{x2x+1}dx$

{.................($I_1$)................} {...............($I_2$)..........}

For$I_1\to$take $\sin =u\Rightarrow \cos dx=du$

$\therefore I_1=\int u^2(1-u^2)du=\int u^{2}du-\int u^{4}du$

$\Rightarrow I_1=\frac{{\sin }^3}{3}-\frac{{\sin }^5}{5}+c_1$

$I_2\to$ take $2x+1=t^2\Rightarrow 2dx=tdt$

$\Rightarrow I_2=\int \frac{(t^2-1)}{2}.t.tdt=\frac{1}{2}\int (t^4-t^2)dt$

$\Rightarrow I_2=\frac{1}{2}[\frac{(2x+1{)}^{5/2}}{5}-\frac{(2x+1{)}^{3/2}}{3}+c_2$

$\therefore y=I_1+I_2$