Question Number 5048 by LMTV last updated on 05/Apr/16
$$\int\left({x}\sqrt{{x}}+\mathrm{1}\right)^{\mathrm{2}} {dx}=? \\ $$$$\int{y}\left(\mathrm{2}{y}+\mathrm{1}\right)^{\mathrm{2}} {dy}=? \\ $$$$\int\frac{\mathrm{1}}{\mathrm{4sec}\:{x}}{dx}=? \\ $$$${y}=\mathrm{sec}^{\mathrm{3}} \:\left(\mathrm{2}{x}+\mathrm{5}\right) \\ $$$${y}'=? \\ $$
Commented by prakash jain last updated on 05/Apr/16
![y=sec^3 (2x+5) Successive application of Chain rule (dy/dx)=(dy/du)∙(du/dx) y′=3sec^2 (2x+5)[sec (2x+5)tan (2x+5)][2] =6sec^3 (2x+5)tan (2x+5) ∫(1/(4sec x))dx=(1/4)∫cos x dx=(1/4)sin x+C](Q5049.png)
$${y}=\mathrm{sec}^{\mathrm{3}} \left(\mathrm{2}{x}+\mathrm{5}\right) \\ $$$$\mathrm{Successive}\:\mathrm{application}\:\mathrm{of}\:\mathrm{Chain}\:\mathrm{rule}\:\frac{{dy}}{{dx}}=\frac{{dy}}{{du}}\centerdot\frac{{du}}{{dx}} \\ $$$${y}'=\mathrm{3sec}^{\mathrm{2}} \left(\mathrm{2}{x}+\mathrm{5}\right)\left[\mathrm{sec}\:\left(\mathrm{2}{x}+\mathrm{5}\right)\mathrm{tan}\:\left(\mathrm{2}{x}+\mathrm{5}\right)\right]\left[\mathrm{2}\right] \\ $$$$=\mathrm{6sec}^{\mathrm{3}} \left(\mathrm{2}{x}+\mathrm{5}\right)\mathrm{tan}\:\left(\mathrm{2}{x}+\mathrm{5}\right) \\ $$$$\int\frac{\mathrm{1}}{\mathrm{4sec}\:{x}}\mathrm{d}{x}=\frac{\mathrm{1}}{\mathrm{4}}\int\mathrm{cos}\:{x}\:\mathrm{d}{x}=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:{x}+{C} \\ $$
Answered by prakash jain last updated on 05/Apr/16
$$\int{y}\left(\mathrm{2}{y}+\mathrm{1}\right)^{\mathrm{2}} {dy}=\int{y}\left(\mathrm{4}{y}^{\mathrm{2}} +\mathrm{4}{y}+\mathrm{1}\right){dy} \\ $$$$=\int\left(\mathrm{4}{y}^{\mathrm{3}} +\mathrm{4}{y}^{\mathrm{2}} +{y}\right){dy} \\ $$$$=\mathrm{4}\frac{{y}^{\mathrm{4}} }{\mathrm{4}}+\mathrm{4}\frac{{y}^{\mathrm{3}} }{\mathrm{3}}+\frac{{y}^{\mathrm{2}} }{\mathrm{2}}+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\left(\mathrm{6}{y}^{\mathrm{4}} +\mathrm{8}{y}^{\mathrm{3}} +\mathrm{3}{y}^{\mathrm{2}} \right)+{C} \\ $$