What is the integral from 0 to pi/4 of 9tan^3(x) ?

The integral from 0 to pi/4 of 9tan^3(x) is 9(-1/2 ln(2)+1/2)

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integral from 0 to pi/4 of 9tan^3(x)

\int_{0}^{\frac{π}{4}} 9 tan^{3} (x) d x

9 (- \frac{1}{2} ln (2) + \frac{1}{2})

Decimal Notation

1.38083 \dots

Solution steps

\int_{0}^{\frac{π}{4}} 9 tan^{3} (x) d x

Take the constant out:

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

= 9 \cdot \int_{0}^{\frac{π}{4}} tan^{3} (x) d x

Rewrite using trig identities

= 9 \cdot \int_{0}^{\frac{π}{4}} (- 1 + sec^{2} (x)) tan (x) d x

Apply u-substitution

: \int_{1}^{2} \frac{- 1 + u ^{2}}{u} d u

= 9 \cdot \int_{1}^{2} \frac{- 1 + u ^{2}}{u} d u

Expand

\frac{- 1 + u ^{2}}{u} : - \frac{1}{u} + u

= 9 \cdot \int_{1}^{2} - \frac{1}{u} + u 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

= 9 (- \int_{1}^{2} \frac{1}{u} d u + \int_{1}^{2} u d u)

\int_{1}^{2} \frac{1}{u} d u = \frac{1}{2} ln (2)

\int_{1}^{2} u d u = \frac{1}{2}

= 9 (- \frac{1}{2} ln (2) + \frac{1}{2})

\frac{d}{d x} (\frac{5 x ^{4} + 3 x ^{3}}{4 x - 9})

e^{x} sin (x)

\int \frac{5 - e ^{x}}{e ^{2 x}} d x

\int \frac{1}{( 64 - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{220 x}{( x + 2 ) ^{2}}

What is the integral from 0 to pi/4 of 9tan^3(x) ?
The integral from 0 to pi/4 of 9tan^3(x) is 9(-1/2 ln(2)+1/2)