Evaluate the integrals

Question

1) ∫x √(x-4) dx

2) ∫(sin(5/x))/(x^2) dx

3) ∫tan^3(4x) sec^2(4x) dx

Answer 1

Hi Ana,

1) ∫x √(x-4) dx x - 4 = u then x = 4 + u then dx = du now back to your question

∫ ( 4 + u ) √u du

∫ (4 u^1/2 + u * u^1/2 ) du

∫ ( 4 u^1/2 + u^3/2 ) du now it is easy = 4 * 2/3 u^3/2 + 2/5 u^5/2+ c now plug in u = x - 4

1) = 8/3 (x -4)^3/2 + 2/5 ( x - 4)^5/2 +c

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2) ∫(sin(5/x))/(x^2) dx 5/x = u then take a derivative of both side

-5/x² dx = du both side divided by -5

dx/x² = -du/5 back to your question∫

∫ -sin u du /5 = -1/5 ∫sinu du = -1/5 ( -cos u ) + c = 1/5 cos u now plug in u = 5/x

2) ∫(sin(5/x))/(x^2) dx = 1/5 cos 5/x + c

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3) ∫tan^3(4x) sec^2(4x) dx it is much easy, only you need to remember tan'(u ) = u' sec^2 (u)

tan'(4x) = 4 sec^2(4x)

tan(4x ) = u

4 sec^2(4x) dx = du both side divided by 4

sec^2(4x) dx = 1/4 du

3) ∫tan^3(4x) sec^2(4x) dx = 1/4∫ u³ du = 1/4 * 1/4 u^4 + c = 1/4 * 1/4 tan⁴(4x) + c = 1/16 tan⁴(4x) + c

I hope it is useful,

Minoo