Solutions to Problem Set 3

Math 211-03

9-7-2017

[Trig powers]

1. Compute $\displaystyle \int
   (\sin x)^8 \cos x\,dx$ .

$$\int (\sin x)^8 \cos x\,dx = \int u^8 (\cos x) \cdot \dfrac{du}{\cos x} = \int u^8\,du = \dfrac{1}{9} u^9 + C = \dfrac{1}{9} (\sin x)^9 + c.$$

$$\left[u = \sin x, \quad du = \cos x\,dx, \quad dx = \dfrac{du}{\cos x}\right]\quad\halmos$$


2. Compute $\displaystyle \int
   (\sin 4 x)^6 (\cos 4 x)^3\,dx$ .

$$\int (\sin 4 x)^6 (\cos 4 x)^3\,dx = \int (\sin 4 x)^6 (\cos 4 x)^2 (\cos 4 x)\,dx = \int (\sin 4 x)^6 [1 - (\sin 4 x)^2] (\cos 4 x)\,dx =$$

$$\left[u = \sin 4 x, \quad du = 4 \cos 4 x\,dx, \quad dx = \dfrac{du}{4 \cos 4 x}\right]$$

$$\int u^6 (1 - u^2) (\cos 4 x) \cdot \dfrac{du}{4 \cos 4 x} = \dfrac{1}{4} \int u^6 (1 - u^2)\,du = \dfrac{1}{4} \int (u^6 - u^8)\,du =$$

$$\dfrac{1}{4} \left(\dfrac{1}{7} u^7 - \dfrac{1}{9} u^9\right) + c = \dfrac{1}{4} \left(\dfrac{1}{7} (\sin 4 x)^7 - \dfrac{1}{9} (\sin 4 x)^9\right) + c.\quad\halmos$$


3. Compute $\displaystyle \int
   (\cos 6 x)^5\,dx$ .

$$\int (\cos 6 x)^5\,dx = \int (\cos 6 x)^4 (\cos 6 x)\,dx = \int [(\cos 6 x)^2]^2 (\cos 6 x)\,dx = \int [1 - (\sin 6 x)^2]^2 (\cos 6 x)\,dx =$$

$$\left[u = \sin 6 x, \quad du = 6 \cos 6 x\,dx, \quad dx = \dfrac{du}{6 \cos 6 x}\right]$$

$$\int (1 - u^2)^2 (\cos 6 x) \cdot \dfrac{du}{6 \cos 6 x} = \dfrac{1}{6} \int (1 - u^2)^2\,du = \dfrac{1}{6} \int (1 - 2 u^2 + u^4)\,du =$$

$$\dfrac{1}{6} \left(u - \dfrac{2}{3} u^3 + \dfrac{1}{5} u^5\right) + c = \dfrac{1}{6} \left(\sin 6 x - \dfrac{2}{3} (\sin 6 x)^3 + \dfrac{1}{5} (\sin 6 x)^5\right) + c.\quad\halmos$$


4. Compute $\displaystyle \int [7
   + (\cos 3 x)^2] (\sin 3 x)^3\,dx$ .

$$\int [7 + (\cos 3 x)^2] (\sin 3 x)^3\,dx = \int [7 + (\cos 3 x)^2] (\sin 3 x)^2 (\sin 3 x)\,dx = \int [7 + (\cos 3 x)^2] [1 - (\cos 3 x)^2] (\sin 3 x)\,dx =$$

$$\left[u = \cos 3 x, \quad du = -3 \sin 3 x\,dx, \quad dx = \dfrac{du}{-3 \sin 3 x}\right]$$

$$\int (7 + u^2)(1 - u^2) (\sin 3 x) \cdot \dfrac{du}{-3 \sin 3 x} = -\dfrac{1}{3} \int (7 + u^2)(1 - u^2)\,du = -\dfrac{1}{3} \int (7 - 6 u^2 - u^4)\,du =$$

$$-\dfrac{1}{3} \left(7 u - 2 u^3 - \dfrac{1}{5} u^5\right) + c = -\dfrac{1}{3} \left(7 \cos 3 x - 2 (\cos 3 x)^3 - \dfrac{1}{5} (\cos 3 x)^5\right) + c.\quad\halmos$$


5. Compute $\displaystyle \int
   (\sin 5 x)^2\,dx$ .

$$\int (\sin 5 x)^2\,dx = \int \dfrac{1}{2} (1 - \cos 10 x)\,dx = \dfrac{1}{2} \left(x - \dfrac{1}{10} \sin 10 x\right) + c.\quad\halmos$$


6. Compute $\displaystyle
   \int_0^{\pi/8} (\cos x)^2 (\sin x)^2\,dx$ .

$$\int_0^{\pi/8} (\cos x)^2 (\sin x)^2\,dx = \int_0^{\pi/8} \dfrac{1}{2} (1 + \cos 2 x) \cdot \dfrac{1}{2} (1 - \cos 2x)\,dx = \dfrac{1}{4} \int_0^{\pi/8} [1 - (\cos 2 x)^2]\,dx =$$

$$\dfrac{1}{4} \int_0^{\pi/8} (\sin 2 x)^2\,dx = \dfrac{1}{4} \int_0^{\pi/8} \dfrac{1}{2} (1 - \cos 4 x)\,dx = \dfrac{1}{8} \left[x - \dfrac{1}{4} \sin 4 x\right]_0^{\pi/8} = \dfrac{\pi}{64} - \dfrac{1}{32} = 0.01783 \ldots .\quad\halmos$$


7. Compute $\displaystyle \int
   (\tan x)^4 (\sec x)^4\,dx$ .

$$\int (\tan x)^4 (\sec x)^4\,dx = \int (\tan x)^4 (\sec x)^2 (\sec x)^2\,dx = \int (\tan x)^4 \left[1 + (\tan x)^2\right] (\sec x)^2\,dx =$$

$$\left[u = \tan x, \quad du = (\sec x)^2\,dx, \quad dx = \dfrac{du}{(\sec x)^2}\right]$$

$$\int u^4 (1 + u^2) (\sec x)^2 \cdot \dfrac{du}{(\sec x)^2} = \int u^4 (1 + u^2)\,du = \int (u^4 + u^6)\,du = \dfrac{1}{5} u^5 + \dfrac{1}{7} u^7 + c =$$

$$\dfrac{1}{5} (\tan x)^5 + \dfrac{1}{7} (\tan x)^7 + c.\quad\halmos$$


8. Compute $\displaystyle \int
   (\tan x)^3 (\sec x)^3\,dx$ .

$$\int (\tan x)^3 (\sec x)^3\,dx = \int (\tan x)^2 (\sec x)^2 \sec x \tan x\,dx = \int [(\sec x)^2 - 1] (\sec x)^2 \sec x \tan x\,dx =$$

$$\left[u = \sec x, \quad du = \sec x \tan x\,dx, \quad dx = \dfrac{du}{\sec x \tan x}\right]$$

$$\int (u^2 - 1) u^2 \sec x \tan x \cdot \dfrac{du}{\sec x \tan x} = \int (u^2 - 1) u^2\,du = \int (u^4 - u^2)\,du =$$

$$\dfrac{1}{5} u^5 - \dfrac{1}{3} u^3 + c = \dfrac{1}{5} (\sec x)^5 - \dfrac{1}{3} (\sec x)^3 + c.\quad\halmos$$


9. Compute $\displaystyle \int
   (\csc x)^4\,dx$ .

$$\int (\csc x)^4 \,dx = \int (\csc x)^2 (\csc x)^2\,dx = \int [1 + (\cot x)^2] (\csc x)^2\,dx = \int (1 + u^2) (\csc x)^2 \cdot \dfrac{du}{-(\csc x)^2} =$$

$$\left[u = \cot x, \quad du = -(\csc x)^2\,dx, \quad dx = \dfrac{du}{-(\csc x)^2}\right]$$

$$-\int (1 + u^2)\,du = -u - \dfrac{1}{3} u^3 + c = -\cot x - \dfrac{1}{3} (\cot x)^3 + c.\quad\halmos$$


10. Compute $\displaystyle \int
   \dfrac{1}{(\cos x) (\sin x)^2}\,dx$ .

Note: You might need the following formula:

$$\int \sec u\,du = \ln |\sec u + \tan u| + c.$$

$$\int \dfrac{1}{(\cos x) (\sin x)^2}\,dx = \int \dfrac{(\sin x)^2 + (\cos x)^2}{(\cos x) (\sin x)^2}\,dx =$$

$$\int \dfrac{(\sin x)^2}{(\cos x) (\sin x)^2}\,dx + \int \dfrac{(\cos x)^2}{(\cos x) (\sin x)^2}\,dx = \int \dfrac{1}{\cos x}\,dx + \int \dfrac{\cos x}{(\sin x)^2}\,dx =$$

$$\int \sec x\,dx + \int \dfrac{\cos x}{\sin x} \cdot \dfrac{1}{\sin x}\,dx = \int \sec x\,dx + \int \cot x \csc x\,dx = \ln |\sec x + \tan x| - \csc x + c.\quad\halmos$$


11. Compute $\displaystyle \int
   (\tan x)^2 (\sec x)\,dx$ .

Note: You might need the following formula:

$$\int (\sec u)^3\,du = \dfrac{1}{2} \sec u \tan u + \dfrac{1}{2} \ln |\sec u + \tan u| + c.$$

$$\int (\tan x)^2 (\sec x)\,dx = \int [(\sec x)^2 - 1] (\sec x)\,dx = \int \left((\sec x)^3 - \sec x\right)\,dx =$$

$$\dfrac{1}{2} \sec x \tan x + \dfrac{1}{2} \ln |\sec x + \tan x| - \ln |\sec x + \tan x| + c = \dfrac{1}{2} \sec x \tan x - \dfrac{1}{2} \ln |\sec x + \tan x| + c.\quad\halmos$$


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