What is the integral of sin^2(10t)sec^4(10t) ?

The integral of sin^2(10t)sec^4(10t) is 1/30 tan^3(10t)+C

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integral of sin^2(10t)sec^4(10t)

\int sin^{2} (10 t) sec^{4} (10 t) d t

\frac{1}{30} tan^{3} (10 t) + C

Solution steps

\int sin^{2} (10 t) sec^{4} (10 t) d t

Rewrite using trig identities

= \int \frac{sin ^{2} ( 10 t )}{cos ^{4} ( 10 t )} d t

Apply u-substitution

: \int \frac{sin ^{2} ( u )}{10 cos ^{4} ( u )} d u

= \int \frac{sin ^{2} ( u )}{10 cos ^{4} ( u )} d u

Take the constant out:

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

= \frac{1}{10} \cdot \int \frac{sin ^{2} ( u )}{cos ^{4} ( u )} d u

Rewrite using trig identities

= \frac{1}{10} \cdot \int \frac{tan ^{2} ( u )}{cos ^{2} ( u )} d u

Apply u-substitution

: \int v^{2} d v

= \frac{1}{10} \cdot \int v^{2} d v

Apply the Power Rule

: \frac{v ^{3}}{3}

= \frac{1}{10} \cdot \frac{v ^{3}}{3}

Substitute back

= \frac{1}{10} \cdot \frac{tan ^{3} ( 10 t )}{3}

Simplify

\frac{1}{10} \cdot \frac{tan ^{3} ( 10 t )}{3} : \frac{1}{30} tan^{3} (10 t)

= \frac{1}{30} tan^{3} (10 t)

Add a constant to the solution

= \frac{1}{30} tan^{3} (10 t) + C

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What is the integral of sin^2(10t)sec^4(10t) ?
The integral of sin^2(10t)sec^4(10t) is 1/30 tan^3(10t)+C