What is the integral of cos^8(x)tan^3(x) ?

The integral of cos^8(x)tan^3(x) is -(cos^6(x))/6+(cos^8(x))/8+C

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integral of cos^8(x)tan^3(x)

\int cos^{8} (x) tan^{3} (x) d x

- \frac{cos ^{6} ( x )}{6} + \frac{cos ^{8} ( x )}{8} + C

Solution steps

\int cos^{8} (x) tan^{3} (x) d x

Rewrite using trig identities

= \int (1 - cos^{2} (x)) sin (x) cos^{5} (x) d x

Apply u-substitution

: \int - u^{5} (1 - u^{2}) d u

= \int - u^{5} (1 - u^{2}) d u

Expand

- u^{5} (1 - u^{2}) : - u^{5} + u^{7}

= \int - u^{5} + u^{7} 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^{5} d u + \int u^{7} d u

\int u^{5} d u = \frac{u ^{6}}{6}

\int u^{7} d u = \frac{u ^{8}}{8}

= - \frac{u ^{6}}{6} + \frac{u ^{8}}{8}

Substitute back

u = cos (x)

= - \frac{cos ^{6} ( x )}{6} + \frac{cos ^{8} ( x )}{8}

Add a constant to the solution

= - \frac{cos ^{6} ( x )}{6} + \frac{cos ^{8} ( x )}{8} + C

\int θ d θ

\int_{1}^{4} \frac{x}{2} + \frac{1}{2 x} d x

f (x) = e^{\frac{x}{2}}

\int 3 t^{8} d t

\int 4 x^{2} - 12 x + 7 d x

What is the integral of cos^8(x)tan^3(x) ?
The integral of cos^8(x)tan^3(x) is -(cos^6(x))/6+(cos^8(x))/8+C