Calculus Volume 1

592

Chapter 5 | Integration

5.5 EXERCISES 254. Why is u -substitution referred to as change of variable ? 255. 2. If f = g ∘ h , when reversing the chain rule, d dx ( g ∘ h )( x ) = g ′( h ( x ) ⎞ ⎠ h ′( x ), should you take u = g ( x ) or u = h ( x )? In the following exercises, verify each identity using differentiation. Then, using the indicated u -substitution, identify f such that the integral takes the form ∫ f ( u ) du .

267. ⌠

3 dx ; u = x 2 −2 x

⌡ ( x −1) ⎛

⎝ x 2 −2 x ⎞ ⎠

268. ⌠ ⌡ ⎛

2 dx ; u = x 3 – 3 x 2

⎝ x 2 −2 x ⎞ ⎠ ⎛

⎝ x 3 −3 x 2 ⎞ ⎠

269. ∫ cos 3 θdθ ; u = sin θ ( Hint : cos 2 θ =1−sin 2 θ )

270. ∫ sin 3 θdθ ; u =cos θ ( Hint : sin 2 θ =1−cos 2 θ )

256. ∫ x x +1 dx = 2 15

( x +1) 3/2 (3 x −2)+ C ; u = x +1

In the following exercises, use a suitable change of variables to determine the indefinite integral. 271. ∫ x (1− x ) 99 dx 272. ⌠ ⌡ t ⎛ ⎝ 1− t 2 ⎞ ⎠ 10 dt 273. ∫ (11 x −7) −3 dx 274. ∫ (7 x −11) 4 dx 275. ∫ cos 3 θ sin θdθ 276. ∫ sin 7 θ cos θdθ 277. ∫ cos 2 ( πt )sin( πt ) dt 278. ∫ sin 2 x cos 3 xdx ( Hint : sin 2 x +cos 2 x =1) 279. ∫ t sin ⎛ ⎝ t 2 ⎞ ⎠ cos ⎛ ⎝ t 2 ⎞ ⎠ dt 280. ∫ t 2 cos 2 ⎛ ⎝ t 3 ⎞ ⎠ sin ⎛ ⎝ t 3 ⎞ ⎠ dt

257.

For

x >1 : ⌠ ⌡

x 2 x −1

⎛ ⎝ 3 x 2 +4 x +8 ⎞

dx = 2 15

x −1

⎠ + C ; u = x −1

258. ⌠ ⌡ x 4 x 2 +9 dx = 1 12 ⎛

3/2

⎝ 4 x 2 +9 ⎞ ⎠

+ C ; u =4 x 2 +9

259. ⌠ ⌡ 260. ⌠ ⌡

x 4 x 2 +9

x 2 +9+ C ; u =4 x 2 +9

dx = 1

4 4

x (4 x 2 +9) 2

; u =4 x 2 +9

dx = − 1

8(4 x 2 +9)

In the following exercises, find the antiderivative using the indicated substitution.

261. ∫ ( x +1) 4 dx ; u = x +1 262. ∫ ( x −1) 5 dx ; u = x −1 263. ∫ (2 x −3) −7 dx ; u =2 x −3 264. ∫ (3 x −2) −11 dx ; u =3 x −2 265. ⌠ ⌡ x x 2 +1 dx ; u = x 2 +1 266. ⌠ ⌡ x 1− x 2 dx ; u =1− x 2

⌠ ⌡ ⎮ x 2 ⎛

dx

281.

2

⎞ ⎠

⎝ x 3 −3

282. ⌠ ⌡

x 3 1− x 2

dx

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