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Alex73 [517]
3 years ago
6

Need help on this calculus

Mathematics
2 answers:
murzikaleks [220]3 years ago
5 0

The general procedure for solving such integrals is bursting out the odd power of the sin and cos.

I = integral [sin^m(t)+ cos^n(t)] dt

Here since both m and n are odd, I'd say pull out a sin(t)cos(t).

I would leave a cos(t) and convert everything else in terms of sin(t) here. Then substitute u = sin(t) and you are good to go.

You could also do u= cos(t) but then du = -sin(t) dt and I don't like the minus sign.

Hope this helps and stay positive :)

N76 [4]3 years ago
4 0
\displaystyle\int\sin^3t\cos^3t\,\mathrm dt

One thing you could do is to expand either a factor of \sin^2t or \cos^2t, then expand the integrand. I'll do the first.

You have

\sin^2t=1-\cos^2t

which means the integral is equivalent to

\displaystyle\int\sin t(1-\cos^2t)\cos^3t\,\mathrm dt

Substitute u=\cos t, so that \mathrm du=-\sin t\,\mathrm dt. This makes it so that the integral above can be rewritten in terms of u as

\displaystyle-\int(1-u^2)u^3\,\mathrm du=\int(u^5-u^3)\,\mathrm du

Now just use the power rule:

\displaystyle\int(u^5-u^3)\,\mathrm du=\dfrac16u^6-\dfrac14u^4+C

Back-substitute to get the antiderivative back in terms of t:

\dfrac16\cos^6t-\dfrac14\cos^4t+C
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t:

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