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# Calculus

Integration of Functions

# Trigonometric and Hyperbolic Substitutions

Page 1
Problems 1-2
Page 2
Problems 3-9

In this section we consider integrals of the form $${\large\int\normalsize} {R\left( {x,\sqrt {a{x^2} + bx + c} } \right)dx} ,$$ where $$R$$ is a rational function of $$x$$ and the radical $${\sqrt {a{x^2} + bx + c} }.$$

To calculate such an integral, we need first to complete the square in the quadratic expression:
${a{x^2} + bx + c } = {a\left( {{{\left( {x + \frac{b}{{2a}}} \right)}^2} }\right.}+{\left.{ \left( {c – \frac{{{b^2}}}{{4a}}} \right)} \right).}$ Making the substitution $$u = x + \large\frac{b}{{2a}}\normalsize,$$ $$du = dx,$$ we can obtain one of the three possible integrals, depending on the values of the coefficients $$a, b$$ and $$c:$$

1. $${\large\int\normalsize} {R\left( {u,\sqrt {{r^2} – {u^2}} } \right)du}$$
2. $${\large\int\normalsize} {R\left( {u,\sqrt {{r^2} + {u^2}} } \right)du}$$
3. $${\large\int\normalsize} {R\left( {u,\sqrt {{u^2} – {r^2}} } \right)du}$$

Then we use the following trigonometric or hyperbolic substitutions to simplify the integrals:

### 1. Integrals of the form $${\large\int\normalsize} {R\left( {u,\sqrt {{r^2} – {u^2}} } \right)du}$$

Trigonometric substitution:
${u = r\sin t,\;\;}\kern-0.3pt {du = r\cos tdt,\;\;}\kern-0.3pt {\sqrt {{r^2} – {u^2}} = r\cos t,\;\;}\kern-0.3pt {t = \arcsin \left( {\frac{u}{r}} \right).}$

### 2. Integrals of the form $${\large\int\normalsize} {R\left( {u,\sqrt {{r^2} + {u^2}} } \right)du}$$

Trigonometric substitution:
${u = r\tan t,\;\;}\kern-0.3pt {du = r\,{\sec ^2}tdt,\;\;}\kern-0.3pt {\sqrt {{r^2} + {u^2}} = r\sec t,\;\;}\kern-0.3pt {t = \arctan \left( {\frac{u}{r}} \right).}$ Hyperbolic substitution:
${u = r\sinh t,\;\;}\kern-0.3pt {du = r\cosh tdt,\;\;}\kern-0.3pt {\sqrt {{r^2} + {u^2}} = r\cosh t,\;\;}\kern-0.3pt {t = \text{arcsinh} \left( {\frac{u}{r}} \right).}$

### 3. Integrals of the form $${\large\int\normalsize} {R\left( {u,\sqrt {{u^2} – {r^2}} } \right)du}$$

Trigonometric substitution:
${u = r\sec t,\;\;}\kern-0.3pt {du = r\tan t\sec tdt,\;\;}\kern-0.3pt {\sqrt {{u^2} – {r^2}} = r\tan t,\;\;}\kern-0.3pt {t = \arccos \left( {\frac{r}{u}} \right).}$ Hyperbolic substitution:
${u = r\cosh t,\;\;}\kern-0.3pt {du = r\sinh tdt,\;\;}\kern-0.3pt {\sqrt {{u^2} – {r^2}} = r\sinh t,\;\;}\kern-0.3pt {t = \text{arccosh} \left( {\frac{u}{r}} \right).}$

Remarks:

• Instead of the trigonometric substitutions in cases $$1, 2, 3$$ you can use the substitutions
$$x = r\cos t,$$ $$x = r\cot t,$$ $$x = r\csc t,$$ respectively.
• Using the formulas given above, we consider only positive values of the root. For example, in strict writing
${\sqrt {{r^2} – {u^2}} } = {\sqrt {{r^2} – {r^2}{{\cos }^2}t} } = {\sqrt {{r^2}{\sin^2}t} } = {\left| {r\sin t} \right|.}$ We suppose here that $$\left| {r\sin t} \right|$$ $$= r\sin t.$$

## Solved Problems

Click on problem description to see solution.

### ✓Example 1

Evaluate the integral $${\large\int\normalsize} {\large\frac{{\sqrt {{a^2} – {x^2}} dx}}{{{x^2}}}\normalsize} .$$

### ✓Example 2

Evaluate the integral $${\large\int\normalsize} {\large\frac{{dx}}{{\sqrt {{{\left( {{a^2} + {x^2}} \right)}^3}} }}\normalsize} .$$

### ✓Example 3

Calculate the integral $${\large\int\normalsize} {\large\frac{{dx}}{{\sqrt {{{\left( {{x^2} – 8} \right)}^3}} }}\normalsize} .$$

### ✓Example 4

Calculate the integral $${\large\int\normalsize} {\large\frac{{dx}}{{\sqrt {\left( {x – a} \right)\left( {b – x} \right)} }}\normalsize} .$$

### ✓Example 5

Evaluate the integral $${\large\int\normalsize} {\large\frac{{\sqrt {{x^2} – {a^2}} }}{x}\normalsize dx} .$$

### ✓Example 6

Evaluate the integral $${\large\int\normalsize} {\sqrt {2x – {x^2}} dx} .$$

### ✓Example 7

Evaluate the integral $${\large\int\normalsize} {\sqrt {9{x^2} – 1} dx} .$$

### ✓Example 8

Calculate the integral $${\large\int\normalsize} {\sqrt {{x^2} + 2x + 3}\,dx} .$$

### ✓Example 9

Evaluate the integral $${\large\int\normalsize} {\large\frac{{\sqrt {{x^2} + 1} }}{x}\normalsize dx} .$$

### Example 1.

Evaluate the integral $${\large\int\normalsize} {\large\frac{{\sqrt {{a^2} – {x^2}} dx}}{{{x^2}}}\normalsize} .$$

#### Solution.

We make the substitution:
${x = a\sin t,\;\;}\kern-0.3pt {dx = a\cos tdt,\;\;}\kern0pt {t = \arcsin \frac{x}{a}.}$ Then
${\int {\frac{{\sqrt {{a^2} – {x^2}} dx}}{{{x^2}}}} } = {\int {\frac{{\sqrt {{a^2} – {a^2}{{\sin }^2}t} }}{{{a^2}{{\sin }^2}t}}a\cos tdt} } = {\int {\frac{{a\cos t}}{{{a^2}{{\sin }^2}t}}a\cos tdt} } = {\int {{{\cot }^2}tdt} = \int {\left( {{{\csc }^2}t – 1} \right)dt} } = { – \cot t – t + C }={{ – \frac{{\sqrt {1 – {{\sin }^2}t} }}{{\sin t}} – t }+{ C }} = {{ – \frac{{\sqrt {1 – \frac{{{x^2}}}{{{a^2}}}} }}{{\frac{x}{a}}} – \arcsin \frac{x}{a} }+{ C }} = {{ – \frac{{\sqrt {{a^2} – {x^2}} }}{x} – \arcsin \frac{x}{a} }+{ C.}}$ To simplify the integral, we used here the trigonometric formula $${\cot ^2}t = {\csc ^2}t – 1.$$

### Example 2.

Evaluate the integral $${\large\int\normalsize} {\large\frac{{dx}}{{\sqrt {{{\left( {{a^2} + {x^2}} \right)}^3}} }}\normalsize} .$$

#### Solution.

We make the hyperbolic substitution: $$x = a\sinh t,$$ $$dx = a\cosh tdt.$$ Using the hyperbolic identity $$1 + {\sinh ^2}t$$ $$= {\cosh ^2}t,$$ we can write:
${\int {\frac{{dx}}{{\sqrt {{{\left( {{a^2} + {x^2}} \right)}^3}} }}} } = {\int {\frac{{a\cosh tdt}}{{\sqrt {{{\left( {{a^2} + {a^2}{{\sinh }^2}t} \right)}^3}} }}} } = {\int {\frac{{a\cosh tdt}}{{{a^3}{{\cosh }^3}t}}} } = {\frac{1}{{{a^2}}}\int {{{\text{sech}}^2}tdt} } = {\frac{1}{{{a^2}}}\tanh t + C } = {\frac{1}{{{a^2}}}\frac{{\sinh t}}{{\cosh t}} + C } = {\frac{1}{{{a^2}}}\frac{{\sinh t}}{{\sqrt {1 + {\sinh^2}t} }} + C } = {\frac{1}{{{a^2}}}\frac{{\frac{x}{a}}}{{\sqrt {1 + \frac{{{x^2}}}{{{a^2}}}} }} + C } = {\frac{1}{{{a^2}}}\frac{x}{{\sqrt {{a^2} + {x^2}} }} + C.}$

Page 1
Problems 1-2
Page 2
Problems 3-9