# Master Trigonometric Integration by Substitution Now

## Substitution of Angle Parts

### Example 1

Find $\displaystyle \int{(2x+3) \sin (x^2+3x)}dx$.

\begin{align} \displaystyle \text{Let } u &= x^2+3x \\ \dfrac{du}{dx} &= 2x + 3 \\ du &= (2x+3)dx \\ \int{(2x+3) \sin (x^2+3x)}dx &= \int{\sin u}du \\ &= -\cos u + C \\ &= -\cos {(x^2+3x)} + C \end{align}

## Substitution of $\sin{x}$

\begin{align} \displaystyle u &= \sin{x} \\ \dfrac{du}{dx} &= \cos{x} \\ du &= \cos{x}du \\ \int{\sin^n{x}\cos{x}}dx &= \int{u^n}du \\ &= \dfrac{u^{n+1}}{n+1} + C \\ &= \dfrac{\sin^{n+1}{x}}{n+1} + C \end{align}

### Example 2

Find $\displaystyle \int{\cos^5{x}\sin^4{x}}dx$.

\begin{align} \displaystyle \text{Let } u &= \sin{x} \\ \dfrac{du}{dx} &= \cos{x} \\ du &= \cos{x}dx \\ \int{\cos^5{x}\sin^4{x}}dx &=\int{\cos^4{x}\sin^4{x}\cos{x}}dx \\ &= \int{(\cos^2{x})^2\sin^4{x}\cos{x}}dx \\ &= \int{(1-\sin^2{x})^2\sin^4{x}\cos{x}}dx \\ &= \int{(1-u^2)^2 u^4}du \\ &= \int{(1-2u^2 + u^4)u^4}du \\ &= \int{(u^4-2u^6 + u^8)}du \\ &= \dfrac{u^{4+1}}{4+1}-\dfrac{2u^{6+1}}{6+1} + \dfrac{u^{8+1}}{8+1} +C \\ &= \dfrac{u^{5}}{5}-\dfrac{2u^{7}}{7} + \dfrac{u^{9}}{9} +c \\ &= \dfrac{\sin^5{x}}{5}-\dfrac{2 \sin^7{x}}{7} + \dfrac{\sin^9{x}}{9} +C \end{align}

## Substitution of $\cos{x}$

\begin{align} \displaystyle u &= \cos{x} \\ \dfrac{du}{dx} &= -\sin{x} \\ -du &= \sin{x}du \\ \int{\cos^n{x}\sin{x}}dx &= -\int{u^n}du \\ &= -\dfrac{u^{n+1}}{n+1} + C \\ &= -\dfrac{\cos^{n+1}{x}}{n+1} + C \end{align}

### Example 3

Find $\displaystyle \int{\sin^3{x}}dx$.

\begin{align} \displaystyle \text{Let } u &= \cos{x} \\ \dfrac{du}{dx} &= -\sin{x} \\ -du &= \sin{x}dx \\ \int{\sin^3{x}}dx &= \int{\sin^2{x}\sin{x}}dx \\ &= \int{(1-\cos^2{x})}\sin{x}dx \\ &= \int{(1-u^2)}(-du) \\ &= \int{(u^2-1)}du \\ &= \dfrac{u^{2+1}}{2+1}-u + C \\ &= \dfrac{u^{3}}{3}-u + C \\ &= \dfrac{\cos^3{x}}{3}-\cos{x} + C \end{align}

### Question 1

Find $\displaystyle \int{\sin{x}\cos^2{x}}dx$.

\begin{aligned} \displaystyle \text{Let } u &= \cos{x} \\ \frac{du}{dx} &= -\sin{x} \\ \int{\sin{x}\cos^2{x}}dx &= \int{\sin{x}u^4}\frac{du}{-\sin{x}} \\ &= -\int{u^4}du \\ &= -\frac{1}{5}u^5 + C \\ &= -\frac{1}{5}\cos^5{x} + C \end{aligned}

### Question 2

Find $\displaystyle \int{\cos{x}\sin^2{x}}dx$.

\begin{aligned} \displaystyle \text{Let } u &= \sin{x} \\ \frac{du}{dx} &= \cos{x} \\ \frac{du}{\cos{x}} &= dx \\ \int{\cos{x}\sin^2{x}}dx &= \int{\cos{x}u^2}\frac{du}{\cos{x}} \\ &= \int{u^2}du \\ &= \frac{1}{3}u^3 + C \\ &= \frac{1}{3} \sin^3{x} + C \end{aligned}

### Question 3

Find $\displaystyle \int{\sin^3{x}}dx$.

\begin{aligned} \displaystyle \text{Let } u &= \cos{x} \\ \frac{du}{dx} &= -\sin{x} \\ -du &= \sin{x}dx \\ \int{\sin^3{x}}dx &= \int{\sin^2{x}\sin{x}}dx \\ &= \int{(1-\cos^2{x})\sin{x}}dx \\ &= \int{(1-u^2)(-du)} \\ &= \int{(u^2-1)}du \\ &= \frac{u^3}{3}-u + C \\ &= \frac{1}{3}\cos^3{x}-\cos{x} + C \end{aligned}

### Question 4

Find $\displaystyle \int{\sin^5{x}}dx$.

\begin{aligned} \displaystyle \text{Let } u &= \cos{x} \\ \frac{du}{dx} &= -\sin{x} \\ -du &= \sin{x}dx \\ \int{\sin^5{x}}dx &= \int{(\sin^2{x})^2\sin{x}}dx \\ &= \int{(1-\cos^2{x})^2}(-du) \\ &= \int{(1-2\cos^2{x} + \cos^4{x})}(-du) \\ &= \int{(1-2u^2+u^4)}(-du) \\ &= -u+\frac{2}{3}u^3-\frac{1}{5}u^5 + C \\ &= -\cos{x} + \frac{2}{3}\cos^3{x}-\frac{1}{5} \cos^5{x} + C \end{aligned}

### Question 5

Find $\displaystyle \int{\cos^3{x}}dx$.

\begin{aligned} \displaystyle \text{Let } u &= \sin{x} \\ \frac{du}{dx} &= \cos{x} \\ du &= \cos{x}dx \\ \int{\cos^3{x}}dx &= \int{\cos^2{x}\cos{x}}dx \\ &= \int{(1-\sin^2{x})}du \\ &= \int{(1-u^2)}du \\ &= u-\frac{1}{3}u^3 + C \\ &= \sin{x}-\frac{1}{3}\sin^3{x} + C \end{aligned}

### Question 6

Find $\displaystyle \int{\cos^5{x}}dx$.

\begin{aligned} \displaystyle \text{Let } u &= \sin{x} \\ \frac{du}{dx} &= \cos{x} \\ du &= \cos{x}dx \\ \int{\cos^5{x}}dx &= \int{(\cos^2{x})^2\cos{x}}dx \\ &= \int{(1-\sin^2{x})^2}du \\ &= \int{(1-u^2)^2}du \\ &= \int{(1-2u^2+u^4)}du \\ &= u-\frac{2}{3}u^3 + \frac{1}{5}u^5 + C \\ &=\sin{x}-\frac{2}{3}\sin^3{x} + \frac{1}{5}\sin^5{x} + C \end{aligned}

### Question 7

Find $\displaystyle \int{\sin^3{x}\cos^3{x}}dx$.

\begin{aligned} \displaystyle \text{Let } u &= \cos{x} \\ \frac{du}{dx} &= -\sin{x} \\ -du &= \sin{x}dx \\ \int{\sin^3{x}\cos^3{x}}dx &= \int{\sin^2{x}\cos^3{x}\sin{x}}dx \\ &= \int{(1-\cos^2{x})\cos^3{x}\sin{x}}dx \\ &= \int{(1-u^2)u^3}(-du) \\ &= \int{(-u^3+u^5)}du \\ &= -\frac{1}{4}u^4 + \frac{1}{6}u^6 + C \\ &= -\frac{1}{4}\cos^4{x} + \frac{1}{6}\cos^6{x} + C \end{aligned}

### Question 8

Find $\displaystyle \int{\sin^5{x}\cos^4{x}}dx$.

\begin{aligned} \displaystyle \text{Let } u &= \cos{x} \\ \frac{du}{dx} &= -\sin{x} \\ -du &= \sin{x}dx \\ \int{\sin^5{x}\cos^4{x}}dx &= \int{(1-\cos^2{x})^2\cos^4{x}}(-du) \\ &= -\int{(1-u^2)^2u^4}du \\ &= -\int{(u^4-2u^6+u^8)}du \\ &= -\frac{1}{5}u^5 + \frac{2}{7}u^7-\frac{1}{9}u^9 + C \\ &= -\frac{1}{5}\cos^5{x} + \frac{2}{7}\cos^7{x}-\frac{1}{9}\cos^9{x} + C \end{aligned}

### Question 9

Find $\displaystyle \int{\sec^4{x}}dx$.

\begin{aligned} \displaystyle \text{Let } u &= \tan{x} \\ \frac{du}{dx} &= \sec^2{x} \\ du &= \sec^2{x}dx \\ \int{\sec^4{x}}dx &= \int{\sec^2{x}\sec^2{x}}dx \\ &= \int{(1+\tan^2{x})\sec^2{x}}dx \\ &= \int{(1+u^2)}du \\ &= u + \frac{1}{3}u^3 + C \\ &= \tan{x} + \frac{1}{3}\tan^3{x} + C \end{aligned}

### Question 10

Find $\displaystyle \int{\sec^6{x}}dx$.

\begin{aligned} \displaystyle \text{Let } u &= \tan{x} \\ \frac{du}{dx} &= \sec^2{x} \\ du &= \sec^2{x}dx \\ \int{\sec^6{x}}dx &= \int{(\sec^2{x})^2\sec^2{x}}dx \\ &= \int{(1+\tan^2{x})^2\sec^2{x}}dx \\ &= \int{(1+u^2)^2}du \\ &= \int{(1+2u^2+u^4)} + C \\ &= u+\frac{2}{3}u^3+\frac{1}{5}u^5+C \\ &= \tan{x} + \frac{2}{3}\tan^3{x} + \frac{1}{5}\tan^5{x} + C \end{aligned}

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