What is the integral of (1-cos^2(x))^2sin(x) ?

The integral of (1-cos^2(x))^2sin(x) is -cos(x)+(2cos^3(x))/3-(cos^5(x))/5+C

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integral of (1-cos^2(x))^2sin(x)

\int (1 - cos^{2} (x))^{2} sin (x) d x

- cos (x) + \frac{2 cos ^{3} ( x )}{3} - \frac{cos ^{5} ( x )}{5} + C

Solution steps

\int (1 - cos^{2} (x))^{2} sin (x) d x

Apply u-substitution

: \int - 1 + 2 u^{2} - u^{4} d u

= \int - 1 + 2 u^{2} - u^{4} d u

Apply the Sum Rule:

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

= - \int 1 d u + \int 2 u^{2} d u - \int u^{4} d u

\int 1 d u = u

\int 2 u^{2} d u = \frac{2 u ^{3}}{3}

\int u^{4} d u = \frac{u ^{5}}{5}

= - u + \frac{2 u ^{3}}{3} - \frac{u ^{5}}{5}

Substitute back

u = cos (x)

= - cos (x) + \frac{2 cos ^{3} ( x )}{3} - \frac{cos ^{5} ( x )}{5}

Add a constant to the solution

= - cos (x) + \frac{2 cos ^{3} ( x )}{3} - \frac{cos ^{5} ( x )}{5} + C

s^{2} + 6

f (x) = 8 x^{3} - 2 x^{2} + 2 x, at x = 2

\int_{0}^{t} sin (2 s) d s

\int \frac{2}{10 + 6 x + x ^{2}} d x

\frac{1}{4 x}

What is the integral of (1-cos^2(x))^2sin(x) ?
The integral of (1-cos^2(x))^2sin(x) is -cos(x)+(2cos^3(x))/3-(cos^5(x))/5+C