What is the integral of (sin^2(x))/(cos^6(x)) ?

The integral of (sin^2(x))/(cos^6(x)) is (tan^3(x))/3+(tan^5(x))/5+C

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integral of (sin^2(x))/(cos^6(x))

\int \frac{sin ^{2} ( x )}{cos ^{6} ( x )} d x

\frac{tan ^{3} ( x )}{3} + \frac{tan ^{5} ( x )}{5} + C

Solution steps

\int \frac{sin ^{2} ( x )}{cos ^{6} ( x )} d x

Rewrite using trig identities

= \int \frac{tan ^{2} ( x )}{cos ^{4} ( x )} d x

Use the basic trigonometric identity:

\frac{1}{cos ( x )} = sec (x)

= \int sec^{4} (x) tan^{2} (x) d x

Rewrite using trig identities

= \int (1 + tan^{2} (x)) sec^{2} (x) tan^{2} (x) d x

Apply u-substitution

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

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

Expand

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

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

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

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

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

Substitute back

u = tan (x)

= \frac{tan ^{3} ( x )}{3} + \frac{tan ^{5} ( x )}{5}

Add a constant to the solution

= \frac{tan ^{3} ( x )}{3} + \frac{tan ^{5} ( x )}{5} + C

\int x lo g_{10} (x) d x

x \to \frac{π}{4} lim (sin (x))

f (x) = \frac{x}{3 x + 4 ln ( x )}

\frac{d y}{d x}, y^{2} = 4 x

\frac{\partial}{\partial y} (x y - y^{2})

What is the integral of (sin^2(x))/(cos^6(x)) ?
The integral of (sin^2(x))/(cos^6(x)) is (tan^3(x))/3+(tan^5(x))/5+C