What is the integral of (tan(2x+1))/(cos^2(2x+1)) ?

The integral of (tan(2x+1))/(cos^2(2x+1)) is 1/4 tan^2(2x+1)+C

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integral of (tan(2x+1))/(cos^2(2x+1))

\int \frac{tan ( 2 x + 1 )}{cos ^{2} ( 2 x + 1 )} d x

\frac{1}{4} tan^{2} (2 x + 1) + C

Solution steps

\int \frac{tan ( 2 x + 1 )}{cos ^{2} ( 2 x + 1 )} d x

Rewrite using trig identities

= \int \frac{sin ( 2 x + 1 )}{cos ^{3} ( 2 x + 1 )} d x

Apply u-substitution

: \int \frac{sin ( u )}{2 cos ^{3} ( u )} d u

= \int \frac{sin ( u )}{2 cos ^{3} ( 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{sin ( u )}{cos ^{3} ( u )} d u

Rewrite using trig identities

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

Apply u-substitution

: \int v d v

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

Apply the Power Rule

: \frac{v ^{2}}{2}

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

Substitute back

= \frac{1}{2} \cdot \frac{tan ^{2} ( 2 x + 1 )}{2}

Simplify

\frac{1}{2} \cdot \frac{tan ^{2} ( 2 x + 1 )}{2} : \frac{1}{4} tan^{2} (2 x + 1)

= \frac{1}{4} tan^{2} (2 x + 1)

Add a constant to the solution

= \frac{1}{4} tan^{2} (2 x + 1) + C

\int (ln (2 x)) d x

\int \frac{x arctan ( x )}{( x ^{2} + 1 ) ^{2}} d x

\int \frac{2 x ^{2} - csc ^{2} ( ln ( x ) )}{x} d x

\int sec^{8} (θ) tan (θ) d θ

\int \frac{1}{49 + x ^{2}} d x

What is the integral of (tan(2x+1))/(cos^2(2x+1)) ?
The integral of (tan(2x+1))/(cos^2(2x+1)) is 1/4 tan^2(2x+1)+C