Reduction formula instead of using identities for trigonometric integration?

aleksbooker

New member
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.

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

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.

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

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. 