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

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

Answers (1)

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}

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