Question Number 86132 by M±th+et£s last updated on 27/Mar/20
$$\int{x}^{\mathrm{3}} \:{sin}\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{6}\right)^{\mathrm{5}} \:{dx} \\ $$
Answered by Kunal12588 last updated on 27/Mar/20
![I=∫x^3 sin(2x^2 +6)^5 dx let 2x^2 +6=t ⇒4x dx = dt I=(1/4)∫((t/2)−3) sin t^5 dt =(1/4)[((t/2)−3)∫sin t^5 −(1/2)∫(∫sin t^5 dt)dt] let ∫sin t^5 dt = F(t)+c I=(1/8)(t−6)F(t)+(1/8)(ta−6a)−(1/8)∫(F(t)+b)dt =(1/8)(t−6)F(t)+(a/8)t−(b/8)t−(1/8)∫F(t)dt−(3/4)a =(1/8)(t−6)F(t)+ct−(1/8)∫F(t)dt + C And now it “may be” easy for the persons who knows ∫sin x^5 dx :)](https://www.tinkutara.com/question/Q86155.png)
$${I}=\int{x}^{\mathrm{3}} \:{sin}\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{6}\right)^{\mathrm{5}} \:{dx} \\ $$$${let}\:\:\mathrm{2}{x}^{\mathrm{2}} +\mathrm{6}={t} \\ $$$$\Rightarrow\mathrm{4}{x}\:{dx}\:=\:{dt} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{4}}\int\left(\frac{{t}}{\mathrm{2}}−\mathrm{3}\right)\:{sin}\:{t}^{\mathrm{5}} \:{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left[\left(\frac{{t}}{\mathrm{2}}−\mathrm{3}\right)\int{sin}\:{t}^{\mathrm{5}} −\frac{\mathrm{1}}{\mathrm{2}}\int\left(\int{sin}\:{t}^{\mathrm{5}} {dt}\right){dt}\right] \\ $$$${let}\:\int{sin}\:{t}^{\mathrm{5}} {dt}\:=\:{F}\left({t}\right)+{c} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{8}}\left({t}−\mathrm{6}\right){F}\left({t}\right)+\frac{\mathrm{1}}{\mathrm{8}}\left({ta}−\mathrm{6}{a}\right)−\frac{\mathrm{1}}{\mathrm{8}}\int\left({F}\left({t}\right)+{b}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\left({t}−\mathrm{6}\right){F}\left({t}\right)+\frac{{a}}{\mathrm{8}}{t}−\frac{{b}}{\mathrm{8}}{t}−\frac{\mathrm{1}}{\mathrm{8}}\int{F}\left({t}\right){dt}−\frac{\mathrm{3}}{\mathrm{4}}{a} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\left({t}−\mathrm{6}\right){F}\left({t}\right)+{ct}−\frac{\mathrm{1}}{\mathrm{8}}\int{F}\left({t}\right){dt}\:\:+\:{C} \\ $$$${And}\:{now}\:{it}\:“{may}\:{be}''\:{easy}\:{for}\:{the}\:{persons} \\ $$$$\left.{who}\:{knows}\:\int{sin}\:{x}^{\mathrm{5}} \:{dx}\:\:\::\right) \\ $$
Commented by M±th+et£s last updated on 27/Mar/20
$${thank}\:{you}\:{sir} \\ $$