What is the integral of (sin(4t-1))/(1-sin^2(4t-1)) ?

The integral of (sin(4t-1))/(1-sin^2(4t-1)) is 1/4 sec(4t-1)+C

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integral of (sin(4t-1))/(1-sin^2(4t-1))

\int \frac{sin ( 4 t - 1 )}{1 - sin ^{2} ( 4 t - 1 )} d t

\frac{1}{4} sec (4 t - 1) + C

Solution steps

\int \frac{sin ( 4 t - 1 )}{1 - sin ^{2} ( 4 t - 1 )} d t

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

: \int - \frac{1}{v ^{2}} d v

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

Take the constant out:

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

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

Apply the Power Rule

: - \frac{1}{v}

= \frac{1}{4} (- (- \frac{1}{v}))

Substitute back

= \frac{1}{4} (- (- \frac{1}{cos ( 4 t - 1 )}))

Simplify

\frac{1}{4} (- (- \frac{1}{cos ( 4 t - 1 )})) : \frac{1}{4} sec (4 t - 1)

= \frac{1}{4} sec (4 t - 1)

Add a constant to the solution

= \frac{1}{4} sec (4 t - 1) + C

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What is the integral of (sin(4t-1))/(1-sin^2(4t-1)) ?
The integral of (sin(4t-1))/(1-sin^2(4t-1)) is 1/4 sec(4t-1)+C