Step 1

The given integral is,

\(\int-(\cos^{7}x-5\cos^{5}x-\cos x)\sin x dx\)

Step 2

The above integral can be split into the following integrals.

\(I=\int -\cos^{7}x\sin x dx+\int 5\cos^{5}x\sin x dx+\int \cos x \sin x dx\)

Step 3

The first integral is evaluated using the substitution as follows.

\(\int -\cos^{7}x\sin x dx=-\int -u^{7}du (\text{Use the substitution,} u=\cos x)\)

\(=\frac{u^{8}}{8}+c_{1}\)

\(=\frac{\cos^{8}x}{8}+c_{1}\)

Step 4

The second integral is evaluated using the substitution as follows.

\(\int 5\cos^{5}x\sin x dx = 5\int \cos^{5}x\sin x dx\)

\(=-5\int u^{5}du\) (Use the substitution, \(u = \cos x\))

\(=\frac{-5}{6}u^{6}+c_{2}\)

\(=\frac{-5}{6}\cos^{6}x+c_{2}\)

Step 5

The third integral is evaluated using the substitution as follows.

\(\int \cos x \sin x dx = \int u du\) (Use the substitution, \(u = \sin x\))

\(=\frac{u^{2}}{2}+c_{3}\)

\(=\frac{\sin^{2}x}{2}+c_{3}\)

Step 6

Thus, the integral \(\int\) can be written as follows.

\(I=\frac{\cos^{8}x}{8}+(\frac{-5}{6})\cos^{6}x+\frac{\sin^{2}x}{2}+C\)