What is the integral of (sin(2t+1))/(cos^2(2t+1)) ?

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

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integral of (sin(2t+1))/(cos^2(2t+1))

\int \frac{sin ( 2 t + 1 )}{cos ^{2} ( 2 t + 1 )} d t

\frac{1}{2} sec (2 t + 1) + C

Solution steps

\int \frac{sin ( 2 t + 1 )}{cos ^{2} ( 2 t + 1 )} d t

Apply u-substitution

: \int \frac{tan ( u )}{2 cos ( u )} d u

= \int \frac{tan ( u )}{2 cos ( u )} d u

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

: \int 1 d v

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

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot 1 \cdot v

Substitute back

= \frac{1}{2} \cdot 1 \cdot sec (2 t + 1)

Simplify

= \frac{1}{2} sec (2 t + 1)

Add a constant to the solution

= \frac{1}{2} sec (2 t + 1) + C

y^{^{'}} = \frac{1 - x}{y}

\frac{d}{d x} (- 3 tan (3 x))

\frac{d}{d x} (- a sin (x))

{\frac{( s + 2 ) ^{2}}{s ^{3}}}

x + 1 + x^{2}

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