How to Integrate cos^2(x)sin^3(x) dx?

[karush]

$$\int \cos^2 \left({x}\right)\sin^3 \left({x}\right)dx$$
$$-\int\cos^2 \left({x}\right)\left(1-\cos^2 \left({x}\right)\right)\sin\left({x}\right)dx
=-\int\left(\cos^2 \left({x}\right)-\cos^4\left({x}\right)\right)\sin\left({x}\right)dx $$
$$u=\cos\left({x}\right)\ \ du=-\sin\left({x}\right)dx $$
So
$$-\int\left({u}^{2}-{u}^{4}\right)du$$

So far?

 

[kaliprasad]

$$\int \cos^2 \left({x}\right)\sin^3 \left({x}\right)dx$$
$$-\int\cos^2 \left({x}\right)\left(1-\cos^2 \left({x}\right)\right)\sin\left({x}\right)dx
=-\int\left(\cos^2 \left({x}\right)-\cos^4\left({x}\right)\right)\sin\left({x}\right)dx $$
$$u=\cos\left({x}\right)\ \ du=-\sin\left({x}\right)dx $$
So
$$-\int\left({u}^{2}-{u}^{4}\right)du$$

So far?

you are almost through where is the problem
integral becomes

$-(\frac{u^3}{3} - \frac{u^5}{5}) + C $ and u = $\cos x$ hence the result

 

[karush]

$$-\left(\frac{\cos^3 \left({x}\right)}{3}-\frac{\cos^5\left({x}\right)}{5}\right)
=\left(\frac{-\sin^2 \left({x}\right)}{5}-\frac{2}{15}\right)\cos^3\left({x}\right)+C$$

Done?

 

[Greg]

I can't figure out where that came from.

At least, the one on the left is correct. :)

\(\displaystyle \int\cos^2(x)\sin^3(x)\,dx\)

\(\displaystyle =\int\cos^2(x)(1-\cos^2(x))\sin(x)\,dx\)

\(\displaystyle =\int(\cos^2(x)-\cos^4(x))\sin(x)\,dx\)

\(\displaystyle u=\cos(x)\quad-du=\sin(x)\,dx\)

\(\displaystyle -\int u^2-u^4\,du=\int u^4-u^2\,du=\dfrac{u^5}{5}-\dfrac{u^3}{3}+C\)

\(\displaystyle \int\cos^2(x)\sin^3(x)\,dx=\dfrac{\cos^5(x)}{5}-\dfrac{\cos^3(x)}{3}+C\)

 

[karush]

The answer on the right is the Ti-nspire answer
The last word on all calculations

 

[MarkFL]

The answer on the right is the Ti-nspire answer
The last word on all calculations

If we take Greg's result:

\(\displaystyle I=\frac{\cos^5(x)}{5}-\frac{\cos^3(x)}{3}+C\)

Factor:

\(\displaystyle I=\cos^3(x)\left(\frac{\cos^2(x)}{5}-\frac{1}{3}\right)+C\)

Apply a Pythagorean identity:

\(\displaystyle I=\cos^3(x)\left(\frac{1-\sin^2(x)}{5}-\frac{1}{3}\right)+C\)

\(\displaystyle I=\cos^3(x)\left(\frac{-\sin^2(x)}{5}+\frac{3}{15}-\frac{5}{15}\right)+C\)

\(\displaystyle I=\left(\frac{-\sin^2(x)}{5}-\frac{2}{15}\right)\cos^3(x)+C\)

And we have the less elegant result spat out by the machine. :)

 

[karush]

The Ti-nspire and book answer are rarely the same for integration. However on this one the book had no answer. The calculator seems to want to factor everything and leave answers in sinx and cosx. why? not sure.