MHB Reduction formula instead of using identities for trigonometric integration?

[aleksbooker1]

This is one of the example problems in my book to show how to deal with integrating trigonometric functions to higher powers, by breaking them down into identities.

$$=\int cos^5x dx$$
$$=\int (cos^2x)^2cos^x dx $$
$$=\int (1-sin^2x)^2*d(sin x)$$
$$=\int (1-u^2)^2 du$$
$$=\int 1-2u^2 + u^4 du$$
$$=u-\frac{2}{3}u^3 + \frac{1}{5}u^5$$
$$=sinx-\frac{2sin^3x}{3}+\frac{sin^5x}{5}+C$$

I tried to do the same problem, but with the reduction formula instead. I'm not sure if I did it right, since the result doesn't really look the same.

$$=\frac{\cos^2{x}\sin{x}}{5} + \frac{4}{5}\int \cos^3{x} dx$$
$$=\frac{1}{5}\cos^4{x}\sin{x} + \frac{4}{5}(\frac{1}{3}\cos^2{x}\sin{x} + \frac{2}{3}\int \cos{x})$$
$$=\frac{1}{5}\cos^4{x}\sin{x}+\frac{4}{15}\cos^2{x}\sin{x}+\frac{8}{15}\sin{x}$$
$$=\frac{3\cos^4{x}\sin{x}+4\cos^2{x}\sin{x}+8\sin{x}}{15}$$
$$=\frac{1}{15}\sin{x}(3\cos^4{x}+4\cos^2{x}+8)+C$$

 

[MarkFL]

Use the Pythagorean identity $\cos^2(x)=1-\sin^2(x)$ on your second form, then expand and collect like terms and you will find it is the same as your first form. :D