Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 4 down vote accepted

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:

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

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

share|cite|improve this answer

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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.