M1272HW1 - Math 1272 Lecture Section 30 Homework Solutions...

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Math 1272 – Lecture Section 30 Homework Solutions I Section 7.5 16) x 2 1 x 2 0 2 / 2 dx Once we have worked for a while with the “ u -substitution” method, we can see that it won’t help us with this integrand. If we try to use u = 1 – x 2 , we will get the differential du = 2 x dx . This would be fine if the exponent of x in the numerator were an odd integer, but if we separate out the factor x dx in the numerator here, there is nothing helpful that we can do with the remaining factor of x from the x 2 term. Instead, we need a different technique which can be applied when a factor appears which is the square root of a sum or difference of two squares: trigonometric substitution. We can set up a right triangle with an angle θ , in which the hypotenuse is equal to 1 and the side opposite the angle θ is x ; the remaining side, adjacent to the angle θ , is given by the radical 1 x 2 . This will make x 1 = sin θ and 1 x 2 1 = cos ; this will also give us the new differential dx = cos θ d θ . Since this is a definite integral, we will also need to transform the limits of integration, yielding x : 0 ( 2)/2 θ = arcsin x : 0 π /4 Our transformed definite integral is x 2 1 x 2 0 2 / 2 dx (sin ) 2 c o s 0 π / 4 ( c o s d ) = sin 2 0 / 4 d . This is now a “trigonometric powers” integral: this particular type, with the integrand being the square of sine or cosine, must be solved using the “trick” of employing the “double-angle” identity for cosine: cos 2 = cos 2 sin 2 = (1 sin 2 ) sin 2 = 1 2 sin 2 using the Pythagorean Identity sin 2 = 1 2 (1 cos 2 ) . We make this replacement to produce the much more easily calculated integral sin 2 0 / 4 d = 1 2 (1 cos 2 ) 0 / 4 d = 1 2 ( 1 2 sin 2 ) 0 / 4 (continued)
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= 1 2 ( π 4 1 2 sin [2 4 ]) (0 1 2 sin [2 0 ]) Ρ Σ ΢ Τ Φ Υ = 1 2 ( 4 1 2 sin 2 0 + 0) = 1 2 ( 4 1 2 1) = 8 1 4 . (We could also work out the indefinite integral, transform it back into the anti-derivative function in terms of x , and then evaluate the anti-derivative at the original limits of integration.) 36) sin 4 x cos 3 x dx This integrand presents a challenge because the arguments of the two trigonometric functions in the product are different from one another. But this can be changed into something much simpler to integrate by using another of the trigonometric identities, the “product-to-sum rules” (page 465 or A30 of Stewart); the relevant one for this problem is sin α cos β = 1 2 sin( + ) + sin( ) [ ] . This allows us to re-write the integral as sin 4 x cos 3 x dx = 1 2 sin(4 x + 3 x ) + x 3 x ) [ ] dx = 1 2 sin 7 x + sin x dx = 1 2 ([ 1 7 cos 7 x ] + [ cos x + C = 1 14 cos 7 x 1 2 cos x + C . 38) tan 5 θ sec 3 d 0 / 4 To solve “trigonometric powers” integrals with two different trigonometric functions, we can make use of one of the three forms of the Pythagorean Identity (page A28), which pair sine with cosine, tangent with secant, and cotangent with cosecant. (Any other pairs of different functions can be re-written as powers of sine and cosine.)
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This note was uploaded on 10/05/2011 for the course MATH 1272 taught by Professor Wilson during the Spring '08 term at Minnesota.

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M1272HW1 - Math 1272 Lecture Section 30 Homework Solutions...

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