Evaluate the indefinite integral. ∫x2sin⁡(x3)dx

Question

Evaluate the indefinite integral.  ∫x2sin⁡(x3)dx

Karen Robbins · Accepted Answer

Step 1
To evaluate the below indefinite integral.

\int x^{2} \sin (x^{3}) d x

Step 2
The given indefinite integral can be evaluate using the u-substitution method.
let

u = x^{3}

\Rightarrow d u = 3 x^{2} d x

\Rightarrow x^{2} d x = \frac{d u}{3}

∴ \int x^{2} \sin (x^{3}) d t = \int \sin (u) (\frac{d u}{3})

= \int \frac{\sin (u)}{3} d u

= \frac{1}{3} \cdot \int \sin (u) d u

= \frac{1}{3} (- \cos (u)) + C (∵ \int \sin (u) d u = - \cos (u))

= \frac{1}{3} (- \cos (x^{3})) + C (∵ u = x^{3})

= - \frac{1}{3} \cos (x^{3}) + C

Thus,

\int x^{2} \sin (x^{3}) d x = - \frac{1}{3} \cos (x^{3}) + C

abonirali59 · Accepted Answer

∫x2sin⁡(x3)dx  We bring the expression 3⋅x2 under the differential sign, i.e.:  3⋅x2dx=d(x3),t=x3  Then the original integral can be written as follows:  ∫sin⁡(t)3dt  This is a tabular integral:  ∫sin⁡(t)3dt=−cos⁡(t)3+C  To write down the final answer, it remains to substitute x3 instead of t.  −cos⁡(x3)3+C

Vasquez · Accepted Answer

Given:∫x2∗sin⁡(x3)dx∫sin⁡(t)3dt13∗∫sin⁡(t)dt13∗(−cos⁡(t))13∗(−cos⁡(x3))Simplify−cos⁡(x3)3Add C∈RAnswer:−cos⁡(x3)3+C,C∈R

Evaluate the indefinite integral. \int x^{2}\sin (x^{3})dx

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