What is the integral from-pi/4 to 0 of (tan(x))^2 ?

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

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integral from-pi/4 to 0 of (tan(x))^2

\int_{- \frac{π}{4}}^{0} (tan (x))^{2} d x

1 - \frac{π}{4}

Decimal

0.21460 \dots

Solution steps

\int_{- \frac{π}{4}}^{0} (tan (x))^{2} d x

Rewrite using trig identities

= \int_{- \frac{π}{4}}^{0} \frac{1 - cos ^{2} ( x )}{cos ^{2} ( x )} d x

Expand

\frac{1 - cos ^{2} ( x )}{cos ^{2} ( x )} : \frac{1}{cos ^{2} ( x )} - 1

= \int_{- \frac{π}{4}}^{0} \frac{1}{cos ^{2} ( x )} - 1 d x

Apply the Sum Rule:

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

= \int_{- \frac{π}{4}}^{0} \frac{1}{cos ^{2} ( x )} d x - \int_{- \frac{π}{4}}^{0} 1 d x

\int_{- \frac{π}{4}}^{0} \frac{1}{cos ^{2} ( x )} d x = 1

\int_{- \frac{π}{4}}^{0} 1 d x = \frac{π}{4}

= 1 - \frac{π}{4}

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