Question

1. Evaluate the following indefinite integrals (each worth 3 marks): (a) int frac{x^3}{sqrt{2x^4 + 1}} dx (b) int frac{sin(1 + sqrt{x})}{sqrt{x}} dx (c) int frac{1}{xsqrt{49x^4 - 4}} dx (d) int cos x e^{sin x} dx (e) int frac{1 + ln^3(x)}{x} dx, x > 0 (f) int sec^2 x tan^5 x dx (g) int frac{x}{sqrt[3]{x - 1}} dx (h) int frac{x^3 - 2x^2 + 1}{x - 3} dx 2. Evaluate the following definite integrals (each worth 4 marks): (a) int_0^1 frac{1}{sqrt{25 - 16x^2}} dx (b) int_0^1 frac{x}{25 + 144x^4} dx (c) int_0^{3/5} frac{x + 3}{sqrt{1 - x^2}} dx (d) int_{pi/6}^{pi/4} cos^3(2x) sin(2x) dx (e) int_5^9 frac{4t^3 - 2t^2 + t}{t - 1} dt (f) int_{1/2}^{5/2} frac{2x - 1}{sqrt{4x - 1}} dx

          1. Evaluate the following indefinite integrals (each worth 3 marks):
(a) int frac{x^3}{sqrt{2x^4 + 1}} dx
(b) int frac{sin(1 + sqrt{x})}{sqrt{x}} dx
(c) int frac{1}{xsqrt{49x^4 - 4}} dx
(d) int cos  x  e^{sin  x} dx
(e) int frac{1 + ln^3(x)}{x} dx, x > 0
(f) int sec^2 x  tan^5 x dx
(g) int frac{x}{sqrt[3]{x - 1}} dx
(h) int frac{x^3 - 2x^2 + 1}{x - 3} dx
2. Evaluate the following definite integrals (each worth 4 marks):
(a) int_0^1 frac{1}{sqrt{25 - 16x^2}} dx
(b) int_0^1 frac{x}{25 + 144x^4} dx
(c) int_0^{3/5} frac{x + 3}{sqrt{1 - x^2}} dx
(d) int_{pi/6}^{pi/4} cos^3(2x) sin(2x) dx
(e) int_5^9 frac{4t^3 - 2t^2 + t}{t - 1} dt
(f) int_{1/2}^{5/2} frac{2x - 1}{sqrt{4x - 1}} dx
        
Show more…
1. Evaluate the following indefinite integrals (each worth 3 marks):
(a) int fracx^3sqrt2x^4 + 1 dx
(b) int fracsin(1 + sqrtx)sqrtx dx
(c) int frac1xsqrt49x^4 - 4 dx
(d) int cos  x  e^sin  x dx
(e) int frac1 + ln^3(x)x dx, x > 0
(f) int sec^2 x  tan^5 x dx
(g) int fracxsqrt[3]x - 1 dx
(h) int fracx^3 - 2x^2 + 1x - 3 dx
2. Evaluate the following definite integrals (each worth 4 marks):
(a) int0^1 frac1sqrt25 - 16x^2 dx
(b) int0^1 fracx25 + 144x^4 dx
(c) int0^3/5 fracx + 3sqrt1 - x^2 dx
(d) intpi/6^pi/4 cos^3(2x) sin(2x) dx
(e) int5^9 frac4t^3 - 2t^2 + tt - 1 dt
(f) int1/2^5/2 frac2x - 1sqrt4x - 1 dx

Added by Tracy V.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Evaluate the following indefinite integrals (each worth 3 marks): (a) ∫ x^3 / √(2x^4 + 1) dx (b) ∫ sin(1 + √x) / √x dx (c) ∫ 1 / (x√(49x^4 - 4)) dx (d) ∫ cos x e^sin x dx (e) ∫ (1 + ln^3(x)) / x dx, x > 0 (f) ∫ sec^2 x tan^5 x dx (g) ∫ x / ³√x - 1 dx (h) ∫ (x^3 - 2x^2 + 1) / (x - 3) dx Evaluate the following definite integrals (each worth 4 marks): (a) ∫₀¹ 1 / √(25 - 16x²) dx (b) ∫₀¹ x / (25 + 144x⁴) dx (c) ∫₀³/⁵ (x + 3) / √(1 - x²) dx (d) ∫π/₆π/₄ cos³(2x) sin(2x) dx (e) ∫₅⁹ (4t³ - 2t² + t) / (t - 1) dt (f) ∫₁/₂⁵/² (2x - 1) / √(4x - 1) dx
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Transcript

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00:01 Hi in the given problem we are required to solve the integral so that integral starts here so this is x cube over 2 root over 2 x to the power 4 plus 1 d x now here we'll be using a substitution method so we'll take the substitution to x 2x2 2 2 plus 1 as equal to t so this means if we differentiate both the sides we get 8 x to the power 3 d x is equal to d t or we can say x cube d x is equal to d t over eight now if you see the numerator we have x cubed d x so this can be written as d t integral 1 over 8 root over t now we can also write this as 1 over 8 integral root over uh or we can write t to the power minus half d t so one over root t is written as t to the power minus half now when we integrate this this is one over eight t to the power minus one over two plus one over minus one over two plus one plus integration constant c so this comes out to be so this is half and this is one over four t…
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