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(a)
Let I = ∫ x 1 + 2 x d x I = \displaystyle\int x\sqrt{1 + 2x}\,dx I = ∫ x 1 + 2 x d x
Put 1 + 2 x = t 2 1 + 2x = t^2 1 + 2 x = t 2 2 d x = 2 t d t 2\,dx = 2t\,dt 2 d x = 2 t d t d x = t d t dx = t\,dt d x = t d t x = t 2 − 1 2 x = \frac{t^2 - 1}{2} x = 2 t 2 − 1
Why?
The awkward part is the root. Putting 1 + 2 x = t 2 1 + 2x = t^2 1 + 2 x = t 2 1 + 2 x \sqrt{1 + 2x} 1 + 2 x t t t t t t
Differentiate both sides: d ( 1 + 2 x ) = 2 d x d(1 + 2x) = 2\,dx d ( 1 + 2 x ) = 2 d x d ( t 2 ) = 2 t d t d(t^2) = 2t\,dt d ( t 2 ) = 2 t d t
2 d x = 2 t d t d x = t d t ( divide by 2 ) \begin{aligned} 2\,dx &= 2t\,dt \\ dx &= t\,dt \quad (\text{divide by } 2) \end{aligned} 2 d x d x = 2 t d t = t d t ( divide by 2 ) The factor x x x t t t 1 + 2 x = t 2 1 + 2x = t^2 1 + 2 x = t 2 x = t 2 − 1 2 x = \frac{t^2 - 1}{2} x = 2 t 2 − 1
1 + 2 x = t 2 2 x = t 2 − 1 ( subtract 1 ) x = t 2 − 1 2 ( divide by 2 ) \begin{aligned} 1 + 2x &= t^2 \\ 2x &= t^2 - 1 \quad (\text{subtract } 1) \\ x &= \frac{t^2 - 1}{2} \quad (\text{divide by } 2) \end{aligned} 1 + 2 x 2 x x = t 2 = t 2 − 1 ( subtract 1 ) = 2 t 2 − 1 ( divide by 2 ) I = ∫ t 2 − 1 2 ⋅ t ⋅ t d t = 1 2 ∫ ( t 4 − t 2 ) d t = 1 2 [ t 5 5 − t 3 3 ] + C \begin{aligned} I &= \int\frac{t^2 - 1}{2}\cdot t\cdot t\,dt \\ &= \frac12\int(t^4 - t^2)\,dt \\ &= \frac12\left[\frac{t^5}{5} - \frac{t^3}{3}\right] + C \end{aligned} I = ∫ 2 t 2 − 1 ⋅ t ⋅ t d t = 2 1 ∫ ( t 4 − t 2 ) d t = 2 1 [ 5 t 5 − 3 t 3 ] + C Why?
Each piece is replaced: x → t 2 − 1 2 x \to \frac{t^2 - 1}{2} x → 2 t 2 − 1 1 + 2 x → t \sqrt{1 + 2x} \to t 1 + 2 x → t d x → t d t dx \to t\,dt d x → t d t ( t 2 − 1 ) t 2 = t 4 − t 2 (t^2 - 1)t^2 = t^4 - t^2 ( t 2 − 1 ) t 2 = t 4 − t 2
t 2 − 1 2 ⋅ t ⋅ t = 1 2 ( t 2 − 1 ) t 2 = 1 2 ( t 2 ⋅ t 2 − 1 ⋅ t 2 ) = 1 2 ( t 4 − t 2 ) \begin{aligned} \frac{t^2 - 1}{2}\cdot t\cdot t &= \frac12 (t^2 - 1)t^2 \\ &= \frac12 (t^2 \cdot t^2 - 1 \cdot t^2) \\ &= \frac12 (t^4 - t^2) \end{aligned} 2 t 2 − 1 ⋅ t ⋅ t = 2 1 ( t 2 − 1 ) t 2 = 2 1 ( t 2 ⋅ t 2 − 1 ⋅ t 2 ) = 2 1 ( t 4 − t 2 ) Each power is integrated with the power rule:
∫ t n d t = t n + 1 n + 1 + C \int t^n\,dt = \frac{t^{n + 1}}{n + 1} + C ∫ t n d t = n + 1 t n + 1 + C ∫ t 4 d t = t 4 + 1 4 + 1 = t 5 5 ∫ t 2 d t = t 2 + 1 2 + 1 = t 3 3 \begin{aligned} \int t^4\,dt &= \frac{t^{4 + 1}}{4 + 1} = \frac{t^5}{5} \\ \int t^2\,dt &= \frac{t^{2 + 1}}{2 + 1} = \frac{t^3}{3} \end{aligned} ∫ t 4 d t ∫ t 2 d t = 4 + 1 t 4 + 1 = 5 t 5 = 2 + 1 t 2 + 1 = 3 t 3 I = t 5 10 − t 3 6 + C I = \frac{t^5}{10} - \frac{t^3}{6} + C I = 10 t 5 − 6 t 3 + C Why?
Multiply the 1 2 \frac12 2 1
1 2 × t 5 5 = t 5 2 × 5 = t 5 10 1 2 × t 3 3 = t 3 2 × 3 = t 3 6 \begin{aligned} \frac12 \times \frac{t^5}{5} &= \frac{t^5}{2 \times 5} = \frac{t^5}{10} \\ \frac12 \times \frac{t^3}{3} &= \frac{t^3}{2 \times 3} = \frac{t^3}{6} \end{aligned} 2 1 × 5 t 5 2 1 × 3 t 3 = 2 × 5 t 5 = 10 t 5 = 2 × 3 t 3 = 6 t 3 Putting back t = 1 + 2 x t = \sqrt{1 + 2x} t = 1 + 2 x
Why?
The answer must be in x x x t = ( 1 + 2 x ) 1 / 2 t = (1 + 2x)^{1/2} t = ( 1 + 2 x ) 1/2 t 5 = ( 1 + 2 x ) 5 / 2 t^5 = (1 + 2x)^{5/2} t 5 = ( 1 + 2 x ) 5/2 t 3 = ( 1 + 2 x ) 3 / 2 t^3 = (1 + 2x)^{3/2} t 3 = ( 1 + 2 x ) 3/2
∴ I = ( 1 + 2 x ) 5 / 2 10 \therefore\ I = \frac{(1 + 2x)^{5/2}}{10} ∴ I = 10 ( 1 + 2 x ) 5/2 − ( 1 + 2 x ) 3 / 2 6 + C - \frac{(1 + 2x)^{3/2}}{6} + C − 6 ( 1 + 2 x ) 3/2 + C
(b)
Let I = ∫ 0 π 2 / 4 sin x x d x I = \displaystyle\int_0^{\pi^2/4}\frac{\sin\sqrt x}{\sqrt x}\,dx I = ∫ 0 π 2 /4 x sin x d x
Put x = t \sqrt x = t x = t x = t 2 x = t^2 x = t 2 d x = 2 t d t dx = 2t\,dt d x = 2 t d t
Why?
x \sqrt x x d x = 2 t d t dx = 2t\,dt d x = 2 t d t t t t
When x = 0 x = 0 x = 0 t = 0 t = 0 t = 0 x = π 2 4 x = \frac{\pi^2}{4} x = 4 π 2 t = π 2 t = \frac{\pi}{2} t = 2 π
Why?
The limits are values of x x x t = x t = \sqrt x t = x 0 = 0 \sqrt0 = 0 0 = 0 π 2 / 4 = π / 2 \sqrt{\pi^2/4} = \pi/2 π 2 /4 = π /2
Watch out: with the new limits we stay in t t t x x x
I = ∫ 0 π / 2 sin t t ⋅ 2 t d t = 2 ∫ 0 π / 2 sin t d t = 2 [ − cos t ] 0 π / 2 = 2 ( − cos π 2 + cos 0 ) = 2 ( 0 + 1 ) \begin{aligned} I &= \int_0^{\pi/2}\frac{\sin t}{t}\cdot 2t\,dt \\ &= 2\int_0^{\pi/2}\sin t\,dt \\ &= 2\Big[-\cos t\Big]_0^{\pi/2} \\ &= 2\left(-\cos\frac{\pi}{2} + \cos 0\right) \\ &= 2(0 + 1) \end{aligned} I = ∫ 0 π /2 t sin t ⋅ 2 t d t = 2 ∫ 0 π /2 sin t d t = 2 [ − cos t ] 0 π /2 = 2 ( − cos 2 π + cos 0 ) = 2 ( 0 + 1 ) Why?
The t t t ∫ sin t d t = − cos t \int\sin t\,dt = -\cos t ∫ sin t d t = − cos t
sin t t ⋅ 2 t = 2 t sin t t = 2 sin t \frac{\sin t}{t}\cdot 2t = \frac{2t\sin t}{t} = 2\sin t t sin t ⋅ 2 t = t 2 t sin t = 2 sin t Upper limit minus lower limit: − cos π 2 − ( − cos 0 ) -\cos\frac{\pi}{2} - (-\cos 0) − cos 2 π − ( − cos 0 ) cos π 2 = 0 \cos\frac{\pi}{2} = 0 cos 2 π = 0 cos 0 = 1 \cos 0 = 1 cos 0 = 1
− cos π 2 − ( − cos 0 ) = − cos π 2 + cos 0 = − 0 + 1 = 1 2 ( 0 + 1 ) = 2 ( 1 ) = 2 \begin{aligned} &-\cos\tfrac{\pi}{2} - (-\cos 0) \\ &= -\cos\tfrac{\pi}{2} + \cos 0 \\ &= -0 + 1 = 1 \\[4pt] &2(0 + 1) = 2(1) = 2 \end{aligned} − cos 2 π − ( − cos 0 ) = − cos 2 π + cos 0 = − 0 + 1 = 1 2 ( 0 + 1 ) = 2 ( 1 ) = 2 ∴ I = 2 \therefore\ I = 2 ∴ I = 2
Where this comes from Chapter 7, Integrals : Integration by substitution.
Question type: Substitution
How to spot it: click to see Type 2. Substitution: when the derivative is hiding in the question
Spot it: a function and its derivative both appear, like log x \log x log x 1 x \frac{1}{x} x 1 x 2 + 1 x^2 + 1 x 2 + 1 x x x
Here: (a) Put 1 + 2 x = t 2 1 + 2x = t^2 1 + 2 x = t 2 t t t d x dx d x t d t t\,dt t d t x x x t 2 − 1 2 \frac{t^2 - 1}{2} 2 t 2 − 1 1 x \frac{1}{\sqrt x} x 1 x \sqrt x x x = t \sqrt x = t x = t 0 0 0 π 2 \frac{\pi}{2} 2 π
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