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

The integral of (sin^2(pix))/(cos^6(pix)) is 1/pi ((tan^3(pix))/3+(tan^5(pix))/5)+C

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

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

\frac{1}{π} (\frac{tan ^{3} ( π x )}{3} + \frac{tan ^{5} ( π x )}{5}) + C

Solution steps

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

Apply u-substitution

: \int \frac{sin ^{2} ( u )}{π cos ^{6} ( u )} d u

= \int \frac{sin ^{2} ( u )}{π cos ^{6} ( u )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{π} \cdot \int \frac{sin ^{2} ( u )}{cos ^{6} ( u )} d u

Rewrite using trig identities

= \frac{1}{π} \cdot \int \frac{tan ^{2} ( u )}{cos ^{4} ( u )} d u

Use the basic trigonometric identity:

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

= \frac{1}{π} \cdot \int sec^{4} (u) tan^{2} (u) d u

Rewrite using trig identities

= \frac{1}{π} \cdot \int (1 + tan^{2} (u)) sec^{2} (u) tan^{2} (u) d u

Apply u-substitution

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

= \frac{1}{π} \cdot \int v^{2} (1 + v^{2}) d v

Expand

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

= \frac{1}{π} \cdot \int v^{2} + v^{4} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{π} (\int v^{2} d v + \int v^{4} d v)

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

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

= \frac{1}{π} (\frac{v ^{3}}{3} + \frac{v ^{5}}{5})

Substitute back

= \frac{1}{π} (\frac{tan ^{3} ( π x )}{3} + \frac{tan ^{5} ( π x )}{5})

Add a constant to the solution

= \frac{1}{π} (\frac{tan ^{3} ( π x )}{3} + \frac{tan ^{5} ( π x )}{5}) + C

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What is the integral of (sin^2(pix))/(cos^6(pix)) ?
The integral of (sin^2(pix))/(cos^6(pix)) is 1/pi ((tan^3(pix))/3+(tan^5(pix))/5)+C