Question

How do I integrate f(x)=sec(x)4−3tan(x)?

Answer 1

Answer:

15ln(tan(x2)+2)−15ln(2tan(x2)−1)+C

Explanation:

∫secx4−3tanx⋅dx

=∫dx4cosx−3sinx

After using y=tan(x2), dx=2dyy2+1, cosx=1−y2y2+1 and sinx=2yy2+1transforms, this integral became

∫2dyy2+14⋅1−y2y2+1−3⋅2yy2+1

=∫2dyy2+14−6y−4y2y2^1

=-∫dy2y2+3y−2

=-∫dy(y+2)⋅(2y−1)

=-15∫5dy(y+2)⋅(2y−1)

=-15∫(2y+4)⋅dy(y+2)(2y−1)+15∫(2y−1)⋅dy(y+2)(2y−1)

=15∫dyy+2−25∫dy2y−1

=15ln(y+2)−15ln(2y−1)+C

=15ln(tan(x2)+2)−15ln(2tan(x2)−1)+C