integration by parts question?

Question

1) integral (4x sec^2 * 2x) dx

2) definite integral from 0 to 1/sqrt(2)
2x * arcsin (x^2) dx


If anyone can help me solve these two problems I'd appreciate it.
I solved all the simpler problems for this HW, but these 2 questions are hard to approach because of trigs( they confuse me)

Thank you

Answer 1

1)

Integral (4x sec^2(2x) dx )

Using integration by parts,

Let u = 4x. dv = sec^2(2x) dx.
du = 4 dx. v = (1/2) tan(2x) dx

4x(1/2)tan(2x) - Integral ( 4(1/2) tan(2x) dx )

Simplifying,

2x tan(2x) - Integral ( 2 tan(2x) dx )

Pulling out the constant from the integral,

2x tan(2x) - 2*Integral (tan(2x) dx )

It's worth remembering that the integral of tan(x) is ln|sec(x)|, which means the integral of tan(2x) is (1/2)ln|sec(2x)|.

2x tan(2x) - 2[ (1/2)ln|sec(2x)|] + C

2x tan(2x) - ln|sec(2x)| + C

2. Integral (0 to 1/sqrt(2), 2x arcsin(x^2) dx )

Here, we use substitution.
Let z = x^2. Then
dz = 2x dx

Our bounds for integration change too; when x = 0, z = 0^2 = 0.
When x = 1/sqrt(2), z = 1/2.
That makes our new bounds 0 to 1/2.

Integral (0 to 1/2, arcsin(z) dz)

Here's where we use integration by parts.

Let u = arcsin(z). dv = dz.
du = 1/sqrt(1 - z^2) dz. v = z.

z arcsin(z) - Integral ( z/sqrt(1 - z^2) dz)

We have to use substitution again.
Let u = 1 - z^2.
du = -2z dz, so
(-1/2) du = z dz

z arcsin(z) - Integral ( 1/sqrt(u) (-1/2) du)
z arcsin(z) + (1/2) Integral ( u^(-1/2) du)

z arcsin(z) + (1/2) [2u^(1/2)]

z arcsin(z) + u^(1/2)

z arcsin(z) + (1 - z^2)^(1/2) {evaluated from 0 to 1/2}

[(1/2)arcsin(1/2) + (1 - (1/2)^2)] - [0 + 1^(1/2)]

[(1/2)(pi/6) + (1 - 1/4)] - 1
[pi/12 + 3/4] - 1

pi/12 + 9/12 - 12/12

(pi - 3)/12