#### Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.27 Question 12

Answer: $I=\frac{e^{x^{3}}}{6}\left ( \cos x^{3}+\sin x^{3} \right )$

Hint: Let $x^{3}=t\Rightarrow 3x^{2}dx=dt$

$\Rightarrow x^{2}dx=\frac{dt}{3}$

Given: $\int x^{2}e^{x^{3}}\cos x^{3}dx$

Solution:

\begin{aligned} &\text { Let } I=x^{2} e^{x^{3}} \cos x^{3} d x \\ &\text { Putting } x^{3}=t \Rightarrow 3 x^{2} d x=d t \\ &\Rightarrow x^{2} d x=\frac{d t}{3} \\ &I=\int e^{t} \cos (t) \times \frac{d t}{3} \\ &I=\frac{1}{3} \int e^{t} \cos (t) d t \\ &3 I=\int e^{t} \cos (t) d t \end{aligned}

As we know that

\begin{aligned} &\int f(x) g(x) d x=f(x) \int g(x) d x-\int\left(f^{\prime}(x) \int g(x) d x\right) d x\\ &\text { Here }\\ &f(t)=\cos t, g(t)=e^{t}\\ &f^{\prime}(t)=-\sin t\\ &\int g(t)=\int e^{t}=e^{t}\\ &3 I=\cos t \times e^{t}-\int(-\sin t) e^{t} d t\\ &3 I=e^{t} \cos t+\int e^{t} \sin t d t \end{aligned}

As we know

\begin{aligned} &\int f(x) g(x) d x=f(x) \int g(x) d x-\int\left(f^{\prime}(x) \int g(x) d x\right) d x\\ &\text { Here }\\ &f(t)=\sin t, g(t)=e^{t}\\ &f^{\prime}(t)=\cos t\\ &\int g(t)=\int e^{t}=e^{t} \end{aligned}

\begin{aligned} &3 I=e^{t} \cos t+\left(\sin t \times t-\int \cos t\right) \times e^{t} d x \\ &3 I=e^{t} \cos t+\left(e^{t} \sin t-\int e^{t} \cos t d t\right) \\ &3 I=e^{t} \cos t+e^{t} \sin t-3 I \\ &3 I+3 I=e^{t} \cos t+e^{t} \sin t \\ &6 I=e^{t} \cos t+e^{t} \sin t \end{aligned}

\begin{aligned} &I=\frac{1}{6} e^{t} \cos t+\frac{1}{6} e^{t} \sin t \\ &I=\frac{e^{t}}{6}(\cos t+\sin t) \\ &I=\frac{e^{x^{3}}}{6}\left(\cos x^{3}+\sin x^{3}\right) \quad\left[\because t=x^{3}\right] \end{aligned}