# Compute the integrals between zero and x

I probably should be embarrassed to ask this question, but I am totally confused on how to compute these integrals. Could someone please help.

$$\int_0^x(3t^2+2t)\sin(t^3+t^2)\,dt$$ $$\int_0^x(3t^2+2t)\sin(x^3+x^2)\,dt$$

Does the first remain the same but replace $t$ with $x$, and with the second one, do I replace the first part with $x$ and find the antiderivative of the second part?

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## 2 Answers

The first integral is nothing special:

$$\int_0^x(3t^2+2t)\sin(t^3+t^2)\,dt$$ Let $u=t^3+t^2\implies du = 3t^2+2t dt$

Thus the integral becomes: $$\int_0^{x^3+x^2}\sin(u)\,du$$ $$\cos u \Bigg|_0^{x^3+x^2} = \cos(x^3+x^2) - 1$$

For the second integral, note that $x$ is constant with respect to the integrand, and can be "pulled out". Thus, we have:

$$\int_0^x(3t^2+2t)\sin(x^3+x^2)\,dt = \sin(x^3+x^2)\int_0^x 3t^2+2t\,dt$$

$$\sin(x^3 + x^2)\left[t^3 + t^2 \Bigg|_0^x\right]$$ $$\sin(x^3 + x^2)\left[x^3 + x^2\right]$$

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Note that

$$\frac{d}{dt} (t^3 + t^2) = 3 t^2 + 2 t$$

In general,

$$\int dt \: f[g(t)] g'(t) = \int dx \: f(x)$$

so that the first integral is

$$\int_0^x dt \: (3 t^2 + 2 t) \sin{(t^3+t^2)} = \int_0^{x^3+x^2} du \: \sin{u}$$

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