what is 16 x 2 Try x 4 tan theta Then x 4 sec 2theta d theta So int x 3 sqrt x 2 16 dx int 64 tan 3 theta times 4 sec 2theta 4sectheta d theta
How do you integrate int x 3 sqrt 16 x 2 dx using trigonometric substitution Calculus Techniques of Integration Integration by Trigonometric Substitution 1 Answer Dy dx x 2 sqrt 16 x 2 sqrt 16 x 2 This is a product of 2 functions so we use the product rule which states that the derivative of the product of 2 functions is the first function times the derivative of the second plus the second function times the derivative of the first Inside this product rule we will also require the power rule which states that the derivative of a
what is 16 x 2
what is 16 x 2
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What Is 16 Plus 16 YouTube
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If we look at the denominator it looks like a rearrangement of Pythagoras theorem a 2 b 2 c 2 rarrc 2 b 2 a 2 So the length of the hypotenuse of our triangle is 4 and the length of the opposite is x Here I intx 3sqrt 16 x 2 dx intx 2sqrt 16 x 2 xdx Let x 4sinu x 2 16sin 2u 2xdx 32sinucosudu i e xdx 16sinucosudu
sin 1 1 4x C int1 sqrt 16 x 2 dx int1 sqrt 16 1 x 2 16 dx int1 4sqrt 1 x 4 2 dx 1 4int1 sqrt 1 x 4 2 dx Let u 1 4x du dx 1 4 4du dx 1 4int1 Int 0 4 x sqrt 16 x 2 dx 4 For this particular integral we can use a trigonometric substitution If we have the form sqrt a 2 u 2 we can do the following u asintheta du acostheta d theta sqrt a 2 u 2 acostheta If it helps you can use a triangle to to visualize this Since we have the integral int 0 4 x sqrt 16 x 2 dx Let a 4 so u 4sintheta
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Int1 x 2sqrt x 2 16 dx sqrt x 2 16 16x C int1 x 2sqrt x 2 16 dx We will use trigonometric substitution to solve this Let s set up our triangle Now let s write the basic three trigonometric functions for angle alpha sinalpha 4 x cosalpha sqrt x 2 16 x tanalpha 4 sqrt x 2 16 x 4 sinalpha x 2 16 sin 2alpha 1 x 2 sin 2alpha 16 Let s take the The answer is x 2 x 2 4 x 3 2x 2 4x 8 Factorise x 4 16 as it is the difference of 2 squares You then get x 2 4 x 2 4 x 2 x 2 4 is also the difference of 2 squares and will factorise to x 2 x 2 so this gives you x 2 x 2 x 2 4 x 2 The x 2 terms cancel out to give you x 2 x 2 4
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what is 16 x 2 - sin 1 1 4x C int1 sqrt 16 x 2 dx int1 sqrt 16 1 x 2 16 dx int1 4sqrt 1 x 4 2 dx 1 4int1 sqrt 1 x 4 2 dx Let u 1 4x du dx 1 4 4du dx 1 4int1