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

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

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

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

\frac{1}{2} sec (2 χ + 1) + C

Solution steps

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

Rewrite using trig identities

= \int \frac{sin ( 2 χ + 1 )}{( cos ( 2 χ ) cos ( 1 ) - sin ( 2 χ ) sin ( 1 ) ) ^{2}} d χ

Use the Angle Sum identity:

cos (s) cos (t) - sin (s) sin (t) = cos (s + t)

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

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 χ + 1)

Simplify

= \frac{1}{2} sec (2 χ + 1)

Add a constant to the solution

= \frac{1}{2} sec (2 χ + 1) + C

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