Evaluate the following integrals: ^2e^ x^3 ^3dx - Vidyadip

Question

Evaluate the following integrals:
$\int\text{x}^2\text{e}^{\text{x}^3}\cos\text{x}^3\text{dx}$

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Answer

Given integral is,
$\text{I}=\int\text{x}^2\text{e}^{\text{x}^3}\cos(\text{x}^3)\text{dx}$
Let $x^3 = t$
$\Rightarrow3\text{x}^2\text{dx}=\text{dt}$
$\Rightarrow\text{x}^2\text{dx}=\frac{\text{dt}}{3}$
Integral becomes,
$\frac{1}{3}\int\text{e}^\text{t}\cos\text{t dt}$
$=\frac{1}{3}\text{I}\ \dots(1)$
Where, $\text{I}=\int\text{e}^\text{t}\cos\text{t dt}$
$\text{I}=\int\text{e}^\text{t}\cos\text{t dt}$
Considering cos t as first and $e^t$ as second function
$\text{I}=\cos\text{t e}^\text{t}-\int-\sin\text{t e}^\text{t}\text{dt}$
$\Rightarrow\text{I}=\text{e}^\text{t}\cos\text{t}+\int\sin\text{t }\text{e}^\text{t}\text{dt}$
Again considering sin t as first and $e^t $as second function
$\text{I}=\text{e}^\text{t}\cos\text{t}+\sin\text{t }\text{e}^\text{t}-\int\cos\text{t e}^\text{t}\text{dt}$
$\Rightarrow\text{I}=\text{e}^\text{t}\cos\text{t}+\sin\text{t e}^\text{t}-1$
$\Rightarrow2\text{I}=\text{e}^\text{t}(\sin\text{t}+\cos\text{t})$
$\Rightarrow\text{I}=\frac{\text{e}^\text{t}}{2}(\sin\text{t}+\cos\text{t})$
$\therefore\ \int\text{x}^2\text{e}^{\text{x}^3}\cos(\text{x}^3)\text{dx}=\frac{1}{3}\Big[\frac{\text{e}^\text{t}}{2}(\sin\text{t}+\cos\text{t})\Big]+\text{C}$ [From (1)]
$=\frac{\text{e}^{\text{x}^3}}{6}(\sin\text{x}^3+\cos\text{x}^3)+\text{C}$

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