# Thread: Integration by Parts

1. ## Integration by Parts

I need help with 4 of these. I tried them, but I keep getting the wrong answer.

1. Evalute the integral of (sin(8x))^3*(cos(8x))^2*dx. Use C to denote an arbitrary constant.

2. Evalute the integral of (cos(5x))^3*dx. Use C to denote an arbitrary constant.

3. Evalute the integral of csc(3x)dx. Use C to denote an arbitrary constant.

4. Evalute the integral of (2-2sinx)/cosx dx. Use C to denote an arbitrary constant.

2. Originally Posted by asnxbbyx113
4. Evalute the integral of (2-2sinx)/cosx dx. Use C to denote an arbitrary constant.
Hint:

$\int \frac{2-2\sin x}{\cos x} dx = \int \frac{2}{\cos x} - \frac{2\sin x}{\cos x} dx$

$= \int 2\sec x - 2\tan x dx$

3. Originally Posted by asnxbbyx113

3. Evalute the integral of csc(3x)dx. Use C to denote an arbitrary constant.
.
$\int \csc 3x dx$

Let $t=3x$:

$\frac{1}{3} \int \csc t dt$

$=\frac{1}{3} \left( \ln|\csc x - \cot x| + C\right)$

4. Originally Posted by asnxbbyx113
I need help with 4 of these. I tried them, but I keep getting the wrong answer.

1. Evalute the integral of (sin(8x))^3*(cos(8x))^2*dx. Use C to denote an arbitrary constant.

2. Evalute the integral of (cos(5x))^3*dx. Use C to denote an arbitrary constant.
.
$\int \cos^3 x dx = \int \cos x \cos^2 x dx = \int \cos x (1-\sin^2 x) dx$

Let $t=\sin x$ then $t'=\cos x$:

$\int (1-t^2)dt = t - \frac{1}{3}t^3+C$

$\sin x - \frac{1}{3}\sin^3 x +C$

Now to find,

$\int \cos^3 5x dx$

Just use the substitution $u=5x$ und do what I do above.

5. Originally Posted by asnxbbyx113
I need help with 4 of these. I tried them, but I keep getting the wrong answer.

1. Evalute the integral of (sin(8x))^3*(cos(8x))^2*dx. Use C to denote an arbitrary constant.
$\int \sin^3 8x \cos^2 8x dx$

$=\frac{1}{8} \int \sin^3 u \cos^2 u du$

$=\frac{1}{8} \int \sin^2 u \cos^2 u \sin u du$

$=\frac{1}{8} \int (1-\cos^2 u)\cos^2 u \sin u du$

Let $t=\cos u$ then,

$-\frac{1}{8} \int (1-t^2)t^2 dt$

You finish!