Class 12: Indefinite Integrals

$\displaystyle \text{Evaluate the following integrals:}$

$\displaystyle \textbf{Question 1: }~\int e^{ax}\cos bx\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int e^{ax}\cos(bx)\,dx.$
$\displaystyle \text{Considering } \cos(bx) \text{ as the first function and } e^{ax} \text{ as the second function.}$
$\displaystyle I=\cos(bx)\frac{e^{ax}}{a}-\int\left(-\sin(bx)\cdot b\right)\frac{e^{ax}}{a}\,dx.$
$\displaystyle I=\frac{e^{ax}\cos(bx)}{a}+\frac{b}{a}\int e^{ax}\sin(bx)\,dx.$
$\displaystyle I=\frac{e^{ax}\cos(bx)}{a}+\frac{b}{a}I_1 \quad \ldots (1).$
$\displaystyle \text{where } I_1=\int e^{ax}\sin(bx)\,dx.$
$\displaystyle \text{Now, } I_1=\int e^{ax}\sin(bx)\,dx.$
$\displaystyle \text{Considering } \sin(bx) \text{ as the first function and } e^{ax} \text{ as the second function.}$
$\displaystyle I_1=\sin(bx)\frac{e^{ax}}{a}-\int\cos(bx)\cdot b\frac{e^{ax}}{a}\,dx.$
$\displaystyle I_1=\frac{e^{ax}\sin(bx)}{a}-\frac{b}{a}\int e^{ax}\cos(bx)\,dx.$
$\displaystyle I_1=\frac{e^{ax}\sin(bx)}{a}-\frac{b}{a}I \quad \ldots (2).$
$\displaystyle \text{From (1) and (2),}$
$\displaystyle I=\frac{e^{ax}\cos(bx)}{a}+\frac{b}{a}\left[\frac{e^{ax}\sin(bx)}{a}-\frac{b}{a}I\right].$
$\displaystyle I=\frac{e^{ax}\cos(bx)}{a}+\frac{be^{ax}\sin(bx)}{a^2}-\frac{b^2}{a^2}I.$
$\displaystyle I\left(1+\frac{b^2}{a^2}\right)=\frac{e^{ax}}{a^2}\left[a\cos(bx)+b\sin(bx)\right].$
$\displaystyle I=\frac{e^{ax}\left[a\cos(bx)+b\sin(bx)\right]}{a^2+b^2}+C.$

$\displaystyle \textbf{Question 2: }~\int e^{ax}\sin(bx+c)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int e^{ax}\sin(bx+C)\,dx.$
$\displaystyle \text{Considering } \sin(bx+C) \text{ as the first function and } e^{ax} \text{ as the second function.}$
$\displaystyle I=\sin(bx+C)\frac{e^{ax}}{a}-\int\cos(bx+C)\cdot b\frac{e^{ax}}{a}\,dx.$
$\displaystyle I=\frac{e^{ax}\sin(bx+C)}{a}-\frac{b}{a}\int e^{ax}\cos(bx+C)\,dx.$
$\displaystyle I=\frac{e^{ax}\sin(bx+C)}{a}-\frac{b}{a}I_1 \quad \ldots (1).$
$\displaystyle \text{where } I_1=\int e^{ax}\cos(bx+C)\,dx.$
$\displaystyle \text{Now, } I_1=\int e^{ax}\cos(bx+C)\,dx.$
$\displaystyle \text{Considering } \cos(bx+C) \text{ as the first function and } e^{ax} \text{ as the second function.}$
$\displaystyle I_1=\cos(bx+C)\frac{e^{ax}}{a}-\int\left(-\sin(bx+C)\cdot b\right)\frac{e^{ax}}{a}\,dx.$
$\displaystyle I_1=\frac{e^{ax}\cos(bx+C)}{a}+\frac{b}{a}\int e^{ax}\sin(bx+C)\,dx.$
$\displaystyle I_1=\frac{e^{ax}\cos(bx+C)}{a}+\frac{b}{a}I \quad \ldots (2).$
$\displaystyle \text{From (1) and (2),}$
$\displaystyle I=\frac{e^{ax}\sin(bx+C)}{a}-\frac{b}{a}\left[\frac{e^{ax}\cos(bx+C)}{a}+\frac{b}{a}I\right].$
$\displaystyle I=\frac{e^{ax}\sin(bx+C)}{a}-\frac{be^{ax}\cos(bx+C)}{a^2}-\frac{b^2}{a^2}I.$
$\displaystyle I\left(1+\frac{b^2}{a^2}\right)=\frac{e^{ax}}{a^2}\left[a\sin(bx+C)-b\cos(bx+C)\right].$
$\displaystyle I=\frac{e^{ax}\left[a\sin(bx+C)-b\cos(bx+C)\right]}{a^2+b^2}+C.$

$\displaystyle \textbf{Question 3: }~\int \cos(\log x)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int \cos(\log x)\,dx.$
$\displaystyle \text{Let } \log x=t.$
$\displaystyle \Rightarrow x=e^{t}.$
$\displaystyle \Rightarrow dx=e^{t}\,dt.$
$\displaystyle I=\int e^{t}\cos t\,dt.$
$\displaystyle \text{Considering } \cos t \text{ as the first function and } e^{t} \text{ as the second function.}$
$\displaystyle I=\cos t\cdot e^{t}-\int(-\sin t)\cdot e^{t}\,dt.$
$\displaystyle I=e^{t}\cos t+\int e^{t}\sin t\,dt.$
$\displaystyle I=e^{t}\cos t+I_1 \quad \ldots (1).$
$\displaystyle \text{where } I_1=\int e^{t}\sin t\,dt.$
$\displaystyle I_1=\int e^{t}\sin t\,dt.$
$\displaystyle \text{Considering } \sin t \text{ as the first function and } e^{t} \text{ as the second function.}$
$\displaystyle I_1=\sin t\cdot e^{t}-\int \cos t\cdot e^{t}\,dt.$
$\displaystyle I_1=e^{t}\sin t-I \quad \ldots (2).$
$\displaystyle \text{From (1) and (2),}$
$\displaystyle I=e^{t}\cos t+e^{t}\sin t-I.$
$\displaystyle 2I=e^{t}(\sin t+\cos t).$
$\displaystyle I=\frac{e^{t}(\sin t+\cos t)}{2}+C.$
$\displaystyle I=\frac{e^{\log x}\,[\sin(\log x)+\cos(\log x)]}{2}+C.$
$\displaystyle I=\frac{x}{2}\,[\sin(\log x)+\cos(\log x)]+C.$

$\displaystyle \textbf{Question 4: }~\int e^{2x}\cos(3x+4)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int e^{2x}\cos(3x+4)\,dx.$
$\displaystyle \text{Considering } \cos(3x+4) \text{ as the first function and } e^{2x} \text{ as the second function.}$
$\displaystyle I=\cos(3x+4)\frac{e^{2x}}{2}-\int\left(-\sin(3x+4)\cdot 3\right)\frac{e^{2x}}{2}\,dx.$
$\displaystyle I=\frac{e^{2x}\cos(3x+4)}{2}+\frac{3}{2}\int e^{2x}\sin(3x+4)\,dx.$
$\displaystyle I=\frac{e^{2x}\cos(3x+4)}{2}+\frac{3}{2}I_1 \quad \ldots (1).$
$\displaystyle \text{where } I_1=\int e^{2x}\sin(3x+4)\,dx.$
$\displaystyle \text{Now, } I_1=\int e^{2x}\sin(3x+4)\,dx.$
$\displaystyle \text{Considering } \sin(3x+4) \text{ as the first function and } e^{2x} \text{ as the second function.}$
$\displaystyle I_1=\sin(3x+4)\frac{e^{2x}}{2}-\int\cos(3x+4)\cdot 3\frac{e^{2x}}{2}\,dx.$
$\displaystyle I_1=\frac{e^{2x}\sin(3x+4)}{2}-\frac{3}{2}\int e^{2x}\cos(3x+4)\,dx.$
$\displaystyle I_1=\frac{e^{2x}\sin(3x+4)}{2}-\frac{3}{2}I \quad \ldots (2).$
$\displaystyle \text{From (1) and (2),}$
$\displaystyle I=\frac{e^{2x}\cos(3x+4)}{2}+\frac{3}{2}\left[\frac{e^{2x}\sin(3x+4)}{2}-\frac{3}{2}I\right].$
$\displaystyle I=\frac{e^{2x}\cos(3x+4)}{2}+\frac{3e^{2x}\sin(3x+4)}{4}-\frac{9}{4}I.$
$\displaystyle I+\frac{9}{4}I=\frac{e^{2x}}{4}\left[2\cos(3x+4)+3\sin(3x+4)\right].$
$\displaystyle \frac{13}{4}I=\frac{e^{2x}}{4}\left[2\cos(3x+4)+3\sin(3x+4)\right].$
$\displaystyle I=\frac{e^{2x}\left[2\cos(3x+4)+3\sin(3x+4)\right]}{13}+C.$

$\displaystyle \textbf{Question 5: }~\int e^{2x}\sin x\cos x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int e^{2x}\sin x\cos x\,dx.$
$\displaystyle I=\frac{1}{2}\int e^{2x}(2\sin x\cos x)\,dx.$
$\displaystyle \Rightarrow I=\frac{1}{2}\int e^{2x}\sin 2x\,dx.$
$\displaystyle \text{Considering } \sin 2x \text{ as the first function and } e^{2x} \text{ as the second function.}$
$\displaystyle I=\frac{1}{2}\left[\sin 2x\frac{e^{2x}}{2}-\int \cos 2x\cdot 2\frac{e^{2x}}{2}\,dx\right].$
$\displaystyle I=\frac{e^{2x}\sin 2x}{4}-\frac{1}{2}\int e^{2x}\cos 2x\,dx.$
$\displaystyle I=\frac{e^{2x}\sin 2x}{4}-\frac{1}{2}I_1 \quad \ldots (1).$
$\displaystyle \text{where } I_1=\int e^{2x}\cos 2x\,dx.$
$\displaystyle \text{Now, } I_1=\int e^{2x}\cos 2x\,dx.$
$\displaystyle \text{Considering } \cos 2x \text{ as the first function and } e^{2x} \text{ as the second function.}$
$\displaystyle I_1=\cos 2x\frac{e^{2x}}{2}-\int\left(-\sin 2x\cdot 2\right)\frac{e^{2x}}{2}\,dx.$
$\displaystyle I_1=\frac{e^{2x}\cos 2x}{2}+\int e^{2x}\sin 2x\,dx.$
$\displaystyle I_1=\frac{e^{2x}\cos 2x}{2}+2I \quad \ldots (2).$
$\displaystyle \text{From (1) and (2),}$
$\displaystyle I=\frac{e^{2x}\sin 2x}{4}-\frac{1}{2}\left[\frac{e^{2x}\cos 2x}{2}+2I\right].$
$\displaystyle I=\frac{e^{2x}\sin 2x}{4}-\frac{e^{2x}\cos 2x}{4}-I.$
$\displaystyle 2I=\frac{e^{2x}}{4}(\sin 2x-\cos 2x).$
$\displaystyle I=\frac{e^{2x}}{8}(\sin 2x-\cos 2x)+C.$

$\displaystyle \textbf{Question 6: }~\int e^{2x}\sin x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int e^{2x}\sin x\,dx.$
$\displaystyle \text{Considering } \sin x \text{ as the first function and } e^{2x} \text{ as the second function.}$
$\displaystyle I=\sin x\frac{e^{2x}}{2}-\int \cos x\frac{e^{2x}}{2}\,dx.$
$\displaystyle I=\frac{e^{2x}\sin x}{2}-\frac{1}{2}\int e^{2x}\cos x\,dx.$
$\displaystyle I=\frac{e^{2x}\sin x}{2}-\frac{1}{2}\left[\cos x\frac{e^{2x}}{2}-\int(-\sin x)\frac{e^{2x}}{2}\,dx\right].$
$\displaystyle I=\frac{e^{2x}\sin x}{2}-\frac{e^{2x}\cos x}{4}-\frac{1}{4}\int e^{2x}\sin x\,dx.$
$\displaystyle I=\frac{e^{2x}(2\sin x-\cos x)}{4}-\frac{1}{4}I.$
$\displaystyle 5I=\frac{e^{2x}(2\sin x-\cos x)}{1}.$
$\displaystyle I=\frac{e^{2x}(2\sin x-\cos x)}{5}+C.$

$\displaystyle \textbf{Question 7: }~\int e^{2x}\sin(3x+1)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have,}$
$\displaystyle I=\int e^{2x}\sin(3x+1)\,dx.$
$\displaystyle \text{Let the first function be } \sin(3x+1) \text{ and the second function be } e^{2x}.$
$\displaystyle \text{First we find the integral of the second function, i.e., } \int e^{2x}\,dx.$
$\displaystyle \int e^{2x}\,dx=\frac{1}{2}e^{2x}.$
$\displaystyle \text{Now, using integration by parts, we get}$
$\displaystyle I=\sin(3x+1)\int e^{2x}\,dx-\int\left(\frac{d(\sin(3x+1))}{dx}\int e^{2x}\,dx\right)dx.$
$\displaystyle I=\frac{1}{2}e^{2x}\sin(3x+1)-\frac{3}{2}\int e^{2x}\cos(3x+1)\,dx.$
$\displaystyle I=\frac{1}{2}e^{2x}\sin(3x+1)-\frac{3}{2}\left[\cos(3x+1)\int e^{2x}\,dx-\int\left(\frac{d(\cos(3x+1))}{dx}\int e^{2x}\,dx\right)dx\right].$
$\displaystyle I=\frac{1}{2}e^{2x}\sin(3x+1)-\frac{3}{2}\left[\frac{1}{2}e^{2x}\cos(3x+1)+\frac{3}{2}\int e^{2x}\sin(3x+1)\,dx\right].$
$\displaystyle I=\frac{1}{2}e^{2x}\sin(3x+1)-\frac{3}{4}e^{2x}\cos(3x+1)-\frac{9}{4}I.$
$\displaystyle I+\frac{9}{4}I=\frac{1}{2}e^{2x}\sin(3x+1)-\frac{3}{4}e^{2x}\cos(3x+1).$
$\displaystyle \frac{13}{4}I=\frac{e^{2x}}{4}\left[2\sin(3x+1)-3\cos(3x+1)\right].$
$\displaystyle I=\frac{e^{2x}}{13}\left[2\sin(3x+1)-3\cos(3x+1)\right]+C.$
$\displaystyle \text{Hence, } \int e^{2x}\sin(3x+1)\,dx=\frac{e^{2x}}{13}\left[2\sin(3x+1)-3\cos(3x+1)\right]+C.$

$\displaystyle \textbf{Question 8: }~\int e^{x}\sin^2 x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int e^{x}\sin^{2}x\,dx.$
$\displaystyle I=\int e^{x}\left(\frac{1-\cos 2x}{2}\right)\,dx.$
$\displaystyle I=\frac{1}{2}\int e^{x}\,dx-\frac{1}{2}\int e^{x}\cos 2x\,dx.$
$\displaystyle I=\frac{e^{x}}{2}-\frac{1}{2}\int e^{x}\cos 2x\,dx \quad \ldots (1).$
$\displaystyle \text{Let } I_1=\int e^{x}\cos 2x\,dx.$
$\displaystyle \text{Considering } \cos 2x \text{ as the first function and } e^{x} \text{ as the second function.}$
$\displaystyle I_1=\cos 2x\cdot e^{x}-\int(-2\sin 2x)\cdot e^{x}\,dx.$
$\displaystyle I_1=e^{x}\cos 2x+2\int e^{x}\sin 2x\,dx.$
$\displaystyle I_1=e^{x}\cos 2x+2\left[\sin 2x\cdot e^{x}-\int(2\cos 2x)\cdot e^{x}\,dx\right].$
$\displaystyle I_1=e^{x}\cos 2x+2e^{x}\sin 2x-4I_1.$
$\displaystyle 5I_1=e^{x}(\cos 2x+2\sin 2x).$
$\displaystyle I_1=\frac{e^{x}}{5}(\cos 2x+2\sin 2x)+C \quad \ldots (2).$
$\displaystyle \text{From (1) and (2),}$
$\displaystyle I=\frac{e^{x}}{2}-\frac{e^{x}}{10}(\cos 2x+2\sin 2x)+C.$

$\displaystyle \textbf{Question 9: }~\int \frac{1}{x^3}\sin(\log x)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int \frac{1}{x^{3}}\sin(\log x)\,dx.$
$\displaystyle \text{Putting } \log x=t.$
$\displaystyle \Rightarrow x=e^{t}.$
$\displaystyle \Rightarrow dx=e^{t}\,dt.$
$\displaystyle \therefore I=\int \frac{1}{e^{3t}}\sin t\cdot e^{t}\,dt.$
$\displaystyle I=\int e^{-2t}\sin t\,dt.$
$\displaystyle \text{Considering } \sin t \text{ as the first function and } e^{-2t} \text{ as the second function.}$
$\displaystyle I=\sin t\left(\frac{e^{-2t}}{-2}\right)-\int \cos t\left(\frac{e^{-2t}}{-2}\right)\,dt.$
$\displaystyle I=-\frac{e^{-2t}\sin t}{2}+\frac{1}{2}\int e^{-2t}\cos t\,dt.$
$\displaystyle I=-\frac{e^{-2t}\sin t}{2}+\frac{1}{2}\left[\cos t\left(\frac{e^{-2t}}{-2}\right)-\int(-\sin t)\left(\frac{e^{-2t}}{-2}\right)\,dt\right].$
$\displaystyle I=-\frac{e^{-2t}\sin t}{2}-\frac{e^{-2t}\cos t}{4}-\frac{1}{4}\int e^{-2t}\sin t\,dt.$
$\displaystyle I=-\frac{e^{-2t}(2\sin t+\cos t)}{4}-\frac{1}{4}I.$
$\displaystyle \frac{5}{4}I=-\frac{e^{-2t}(2\sin t+\cos t)}{4}.$
$\displaystyle I=-\frac{e^{-2t}}{5}(2\sin t+\cos t)+C.$
$\displaystyle I=-\frac{e^{-2\log x}}{5}[2\sin(\log x)+\cos(\log x)]+C.$
$\displaystyle I=-\frac{1}{5x^{2}}[2\sin(\log x)+\cos(\log x)]+C.$

$\displaystyle \textbf{Question 10: }~\int e^{2x}\cos^2 x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int e^{2x}\cos^{2}x\,dx.$
$\displaystyle I=\int e^{2x}\left(\frac{1+\cos 2x}{2}\right)\,dx.$
$\displaystyle I=\frac{1}{2}\int e^{2x}\,dx+\frac{1}{2}\int e^{2x}\cos 2x\,dx.$
$\displaystyle I=\frac{e^{2x}}{4}+\frac{1}{2}I_1 \quad \ldots (1).$
$\displaystyle \text{where } I_1=\int e^{2x}\cos 2x\,dx.$
$\displaystyle \text{Considering } \cos 2x \text{ as the first function and } e^{2x} \text{ as the second function.}$
$\displaystyle I_1=\cos 2x\frac{e^{2x}}{2}-\int(-2\sin 2x)\frac{e^{2x}}{2}\,dx.$
$\displaystyle I_1=\frac{e^{2x}\cos 2x}{2}+\int e^{2x}\sin 2x\,dx.$
$\displaystyle \text{Considering } \sin 2x \text{ as the first function and } e^{2x} \text{ as the second function.}$
$\displaystyle I_1=\frac{e^{2x}\cos 2x}{2}+\sin 2x\frac{e^{2x}}{2}-\int(2\cos 2x)\frac{e^{2x}}{2}\,dx.$
$\displaystyle I_1=\frac{e^{2x}(\cos 2x+\sin 2x)}{2}-I_1.$
$\displaystyle 2I_1=\frac{e^{2x}(\cos 2x+\sin 2x)}{2}.$
$\displaystyle I_1=\frac{e^{2x}(\cos 2x+\sin 2x)}{4} \quad \ldots (2).$
$\displaystyle \text{From (1) and (2),}$
$\displaystyle I=\frac{e^{2x}}{4}+\frac{e^{2x}}{8}(\cos 2x+\sin 2x)+C.$

$\displaystyle \textbf{Question 11: }~\int e^{-2x}\sin x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int e^{-2x}\sin x\,dx.$
$\displaystyle \text{Considering } \sin x \text{ as the first function and } e^{-2x} \text{ as the second function.}$
$\displaystyle I=\sin x\left(\frac{e^{-2x}}{-2}\right)-\int \cos x\left(\frac{e^{-2x}}{-2}\right)\,dx.$
$\displaystyle I=-\frac{e^{-2x}\sin x}{2}+\frac{1}{2}\int e^{-2x}\cos x\,dx.$
$\displaystyle I=-\frac{e^{-2x}\sin x}{2}+\frac{1}{2}I_1 \quad \ldots (1).$
$\displaystyle \text{where } I_1=\int e^{-2x}\cos x\,dx.$
$\displaystyle \text{Considering } \cos x \text{ as the first function and } e^{-2x} \text{ as the second function.}$
$\displaystyle I_1=\cos x\left(\frac{e^{-2x}}{-2}\right)-\int(-\sin x)\left(\frac{e^{-2x}}{-2}\right)\,dx.$
$\displaystyle I_1=-\frac{e^{-2x}\cos x}{2}-\frac{1}{2}\int e^{-2x}\sin x\,dx.$
$\displaystyle I_1=-\frac{e^{-2x}\cos x}{2}-\frac{1}{2}I \quad \ldots (2).$
$\displaystyle \text{From (1) and (2),}$
$\displaystyle I=-\frac{e^{-2x}\sin x}{2}+\frac{1}{2}\left[-\frac{e^{-2x}\cos x}{2}-\frac{1}{2}I\right].$
$\displaystyle I=-\frac{e^{-2x}\sin x}{2}-\frac{e^{-2x}\cos x}{4}-\frac{1}{4}I.$
$\displaystyle I+\frac{1}{4}I=-\frac{e^{-2x}(2\sin x+\cos x)}{4}.$
$\displaystyle \frac{5}{4}I=-\frac{e^{-2x}(2\sin x+\cos x)}{4}.$
$\displaystyle I=\frac{e^{-2x}}{5}(-2\sin x-\cos x)+C.$

$\displaystyle \textbf{Question 12: }~\int x^2 e^{x^3}\cos x^3\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given integral is,}$
$\displaystyle \int x^{2}e^{x^{3}}\cos(x^{3})\,dx.$
$\displaystyle \text{Let } x^{3}=t.$
$\displaystyle \Rightarrow 3x^{2}\,dx=dt.$
$\displaystyle \Rightarrow x^{2}\,dx=\frac{dt}{3}.$
$\displaystyle \text{Integral becomes,}$
$\displaystyle \frac{1}{3}\int e^{t}\cos t\,dt.$
$\displaystyle =\frac{1}{3}I \quad \ldots (1).$
$\displaystyle \text{where } I=\int e^{t}\cos t\,dt.$
$\displaystyle I=\int e^{t}\cos t\,dt.$
$\displaystyle \text{Considering } \cos t \text{ as the first function and } e^{t} \text{ as the second function.}$
$\displaystyle I=\cos t\cdot e^{t}-\int(-\sin t)\cdot e^{t}\,dt.$
$\displaystyle I=e^{t}\cos t+\int e^{t}\sin t\,dt.$
$\displaystyle \text{Again considering } \sin t \text{ as the first function and } e^{t} \text{ as the second function.}$
$\displaystyle I=e^{t}\cos t+\sin t\cdot e^{t}-\int \cos t\cdot e^{t}\,dt.$
$\displaystyle I=e^{t}\cos t+e^{t}\sin t-I.$
$\displaystyle 2I=e^{t}(\sin t+\cos t).$
$\displaystyle I=\frac{e^{t}}{2}(\sin t+\cos t).$
$\displaystyle \therefore \int x^{2}e^{x^{3}}\cos(x^{3})\,dx=\frac{1}{3}\left[\frac{e^{t}}{2}(\sin t+\cos t)\right]+C \quad [\text{From (1)}].$
$\displaystyle =\frac{e^{x^{3}}}{6}(\sin x^{3}+\cos x^{3})+C.$