Integrals problems please help

Hi
Can any help me with these integrals needed help urgently. Please
Thanks In advance

Consider the following integrals
a) ʃ tan 3x dx + ʃ 2 cos4x dx

b) Find ʃ dx / (x^2 + 1) using the substitution
Giving your final answer in terms of x

Thanks

The first one seems to be very easy:

lets do each part separately, and later combine:

a) ʃ tan 3x dx = ʃ sin(3x)/cos(3x) dx = ʃ -1/3 * d(cos(3x))/cos(3x)= - 1/3 ln(cos(3x))

ʃ 2 cos4x dx = ʃ 2*1/4* d(sin(4x)) = 1/2 * sin(4x)

And combined: ʃ tan 3x dx + ʃ 2 cos4x dx = - 1/3 ln(cos(3x)) + 1/2 * sin(4x)

b) ʃ dx / (x^2 + 1)

let x = tan(t)
dx = 1/cos^2 (t)dt

ʃ dx / (x^2 + 1) = ʃ (1/cos^2 t)dt/(tan^2 t + 1) = ʃ (1/cos^2 t)dt/(1/cos^2 t) = ʃdt = t = arctan(x)

since we assumed that x = tan(t), so t = arctan(x)

For part a), the only slightly tricky part is finding the indefinite integral of tan(x).

ʃ tan(x)dx = ʃ [sin(x)/cos(x)]dx
Let u = cos(x)
Then du = -sin(x)dx
ʃ (-1/u)du
-ln|u| + C --> -ln|cos(x)| + C

The rest is easy enough, with chain rule, etc.

a) (1/3) * ln(cos(x)) + (1/2) * sin(4x) + C
b) arctan(x) + C

How to find the second one?

dx / (1 + x^2)

x = tan(t)
dx = sec(t)^2 * dt

sec(t)^2 * dt / (1 + tan(t)^2) =>
sec(t)^2 * dt / sec(t)^2 =>
dt

Integrate

int(dt) = t + C

x = tan(t)
arctan(x) = t

t + C =>
arctan(x) + C