What is the integral of cos^3(t)-sin^2(t)cos(t) ?

The integral of cos^3(t)-sin^2(t)cos(t) is sin(t)-2/3 sin^3(t)+C

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integral of cos^3(t)-sin^2(t)cos(t)

\int cos^{3} (t) - sin^{2} (t) cos (t) d t

sin (t) - \frac{2}{3} sin^{3} (t) + C

Solution steps

\int cos^{3} (t) - sin^{2} (t) cos (t) d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int cos^{3} (t) d t - \int sin^{2} (t) cos (t) d t

\int cos^{3} (t) d t = sin (t) - \frac{sin ^{3} ( t )}{3}

\int sin^{2} (t) cos (t) d t = \frac{sin ^{3} ( t )}{3}

= sin (t) - \frac{sin ^{3} ( t )}{3} - \frac{sin ^{3} ( t )}{3}

Simplify

sin (t) - \frac{sin ^{3} ( t )}{3} - \frac{sin ^{3} ( t )}{3} : sin (t) - \frac{2}{3} sin^{3} (t)

= sin (t) - \frac{2}{3} sin^{3} (t)

Add a constant to the solution

= sin (t) - \frac{2}{3} sin^{3} (t) + C

y = x^{2} - 5 x - 6

f (x) = \frac{( x + 9 )}{x ^{2} - 7 x + 1}

g (x) = \frac{7}{x}

\frac{s}{( s + 1 ) ^{2}}

\int \frac{4 x ^{5}}{1 - 12 x ^{6}} d x

What is the integral of cos^3(t)-sin^2(t)cos(t) ?
The integral of cos^3(t)-sin^2(t)cos(t) is sin(t)-2/3 sin^3(t)+C