Question

Need help on this calculus

Answer 1

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 :)

Answer 2

$\displaystyle\int\sin^3t\cos^3t\,\mathrm dt$

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 $t$:

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