What is the integral of (1-t^2)^3 ?

The integral of (1-t^2)^3 is t-t^3+(3t^5)/5-(t^7)/7+C

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integral of (1-t^2)^3

integral

(1 - t^{2})^{3}

t - t^{3} + \frac{3 t ^{5}}{5} - \frac{t ^{7}}{7} + C

Solution steps

\int (1 - t^{2})^{3} d t

Expand

(1 - t^{2})^{3} : 1 - 3 t^{2} + 3 t^{4} - t^{6}

= \int 1 - 3 t^{2} + 3 t^{4} - t^{6} 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 1 d t - \int 3 t^{2} d t + \int 3 t^{4} d t - \int t^{6} d t

\int 1 d t = t

\int 3 t^{2} d t = t^{3}

\int 3 t^{4} d t = \frac{3 t ^{5}}{5}

\int t^{6} d t = \frac{t ^{7}}{7}

= t - t^{3} + \frac{3 t ^{5}}{5} - \frac{t ^{7}}{7}

Add a constant to the solution

= t - t^{3} + \frac{3 t ^{5}}{5} - \frac{t ^{7}}{7} + C

What is the integral of (1-t^2)^3 ?
The integral of (1-t^2)^3 is t-t^3+(3t^5)/5-(t^7)/7+C