What is the integral of cos^7(x)tan^3(x) ?

The integral of cos^7(x)tan^3(x) is -(cos^5(x))/5+(cos^7(x))/7+C

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integral of cos^7(x)tan^3(x)

\int cos^{7} (x) tan^{3} (x) d x

- \frac{cos ^{5} ( x )}{5} + \frac{cos ^{7} ( x )}{7} + C

Solution steps

\int cos^{7} (x) tan^{3} (x) d x

Rewrite using trig identities

= \int (1 - cos^{2} (x)) sin (x) cos^{4} (x) d x

Apply u-substitution

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

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

Expand

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

= \int - u^{4} + u^{6} 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^{4} d u + \int u^{6} d u

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

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

= - \frac{u ^{5}}{5} + \frac{u ^{7}}{7}

Substitute back

u = cos (x)

= - \frac{cos ^{5} ( x )}{5} + \frac{cos ^{7} ( x )}{7}

Add a constant to the solution

= - \frac{cos ^{5} ( x )}{5} + \frac{cos ^{7} ( x )}{7} + C

\frac{d}{d x} (x^{5})

\int \frac{8 x}{1 - x ^{4}} d x

\int (cos^{7} (x)) sin (x) d x

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

y^{^{'}} + \frac{y}{x} = 2 x^{4} y^{2}

What is the integral of cos^7(x)tan^3(x) ?
The integral of cos^7(x)tan^3(x) is -(cos^5(x))/5+(cos^7(x))/7+C