What is the integral of cos^9(x)tan^3(x) ?

The integral of cos^9(x)tan^3(x) is -(cos^7(x))/7+(cos^9(x))/9+C

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integral of cos^9(x)tan^3(x)

\int cos^{9} (x) tan^{3} (x) d x

- \frac{cos ^{7} ( x )}{7} + \frac{cos ^{9} ( x )}{9} + C

Solution steps

\int cos^{9} (x) tan^{3} (x) d x

Rewrite using trig identities

= \int (1 - cos^{2} (x)) sin (x) cos^{6} (x) d x

Apply u-substitution

: \int - u^{6} (1 - u^{2}) d u

= \int - u^{6} (1 - u^{2}) d u

Expand

- u^{6} (1 - u^{2}) : - u^{6} + u^{8}

= \int - u^{6} + u^{8} 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 u^{6} d u + \int u^{8} d u

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

\int u^{8} d u = \frac{u ^{9}}{9}

= - \frac{u ^{7}}{7} + \frac{u ^{9}}{9}

Substitute back

u = cos (x)

= - \frac{cos ^{7} ( x )}{7} + \frac{cos ^{9} ( x )}{9}

Add a constant to the solution

= - \frac{cos ^{7} ( x )}{7} + \frac{cos ^{9} ( x )}{9} + C

25 y^{^{''}} - 40 y^{^{'}} + 16 y = 0, y (0) = 2, y^{^{'}} (0) = 3

\int \frac{24 x ^{2}}{36 - x ^{2}} d x

x \to 1 lim (6 x + 4)

\frac{( e ^{\frac{s}{2}} )}{s ^{2}}

x = y^{2} - 7, x = 1 - y^{2}, - 2 \leq y \leq 2

What is the integral of cos^9(x)tan^3(x) ?
The integral of cos^9(x)tan^3(x) is -(cos^7(x))/7+(cos^9(x))/9+C